#include <bits/stdc++.h>
using namespace std;
using ui = unsigned int;
using ull = unsigned long long;
using ll = long long;
#define endl '\n'
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int maxn = 2e5 + 10;
const int mod = 1000000007;

#include <bits/stdc++.h>
using namespace std;

#define int ll
using point_t = long long; // 全局数据类型，可修改为 long long 等

constexpr point_t eps = 1e-8;
constexpr long double PI = 3.1415926535897932384l;

// 点与向量
template <typename T>
struct point
{
    T x, y;

    bool operator==(const point &a) const { return (abs(x - a.x) <= eps && abs(y - a.y) <= eps); }
    bool operator<(const point &a) const
    {
        if (abs(x - a.x) <= eps)
            return y < a.y - eps;
        return x < a.x - eps;
    }
    bool operator>(const point &a) const { return !(*this < a || *this == a); }
    point operator+(const point &a) const { return {x + a.x, y + a.y}; }
    point operator-(const point &a) const { return {x - a.x, y - a.y}; }
    point operator-() const { return {-x, -y}; }
    point operator*(const T k) const { return {k * x, k * y}; }
    point operator/(const T k) const { return {x / k, y / k}; }
    T operator*(const point &a) const { return x * a.x + y * a.y; } // 点积
    T operator^(const point &a) const { return x * a.y - y * a.x; } // 叉积，注意优先级
    int toleft(const point &a) const
    {
        const auto t = (*this) ^ a;
        return (t > eps) - (t < -eps);
    } // to-left 测试
    T len2() const { return (*this) * (*this); }                  // 向量长度的平方
    T dis2(const point &a) const { return (a - (*this)).len2(); } // 两点距离的平方

    // 涉及浮点数
    long double len() const { return sqrtl(len2()); }                                                                      // 向量长度
    long double dis(const point &a) const { return sqrtl(dis2(a)); }                                                       // 两点距离
    long double ang(const point &a) const { return acosl(max(-1.0l, min(1.0l, ((*this) * a) / (len() * a.len())))); }      // 向量夹角
    point rot(const long double rad) const { return {x * cos(rad) - y * sin(rad), x * sin(rad) + y * cos(rad)}; }          // 逆时针旋转（给定角度）
    point rot(const long double cosr, const long double sinr) const { return {x * cosr - y * sinr, x * sinr + y * cosr}; } // 逆时针旋转（给定角度的正弦与余弦）
};

using Point = point<point_t>;

// 极角排序
struct argcmp
{
    bool operator()(const Point &a, const Point &b) const
    {
        const auto quad = [](const Point &a)
        {
            if (a.y < -eps)
                return 1;
            if (a.y > eps)
                return 4;
            if (a.x < -eps)
                return 5;
            if (a.x > eps)
                return 3;
            return 2;
        };
        const int qa = quad(a), qb = quad(b);
        if (qa != qb)
            return qa < qb;
        const auto t = a ^ b;
        // if (abs(t)<=eps) return a*a<b*b-eps;  // 不同长度的向量需要分开
        return t > eps;
    }
};

// 直线
template <typename T>
struct line
{
    point<T> p, v; // p 为直线上一点，v 为方向向量

    bool operator==(const line &a) const { return v.toleft(a.v) == 0 && v.toleft(p - a.p) == 0; }
    int toleft(const point<T> &a) const { return v.toleft(a - p); } // to-left 测试
    bool operator<(const line &a) const                             // 半平面交算法定义的排序
    {
        if (abs(v ^ a.v) <= eps && v * a.v >= -eps)
            return toleft(a.p) == -1;
        return argcmp()(v, a.v);
    }

    // 涉及浮点数
    point<T> inter(const line &a) const { return p + v * ((a.v ^ (p - a.p)) / (v ^ a.v)); } // 直线交点
    long double dis(const point<T> &a) const { return abs(v ^ (a - p)) / v.len(); }         // 点到直线距离
    point<T> proj(const point<T> &a) const { return p + v * ((v * (a - p)) / (v * v)); }    // 点在直线上的投影
};

using Line = line<point_t>;

// 线段
template <typename T>
struct segment
{
    point<T> a, b;

    bool operator<(const segment &s) const { return make_pair(a, b) < make_pair(s.a, s.b); }

    // 判定性函数建议在整数域使用

    // 判断点是否在线段上
    // -1 点在线段端点 | 0 点不在线段上 | 1 点严格在线段上
    int is_on(const point<T> &p) const
    {
        if (p == a || p == b)
            return -1;
        return (p - a).toleft(p - b) == 0 && (p - a) * (p - b) < -eps;
    }

    // 判断线段直线是否相交
    // -1 直线经过线段端点 | 0 线段和直线不相交 | 1 线段和直线严格相交
    int is_inter(const line<T> &l) const
    {
        if (l.toleft(a) == 0 || l.toleft(b) == 0)
            return -1;
        return l.toleft(a) != l.toleft(b);
    }

    // 判断两线段是否相交
    // -1 在某一线段端点处相交 | 0 两线段不相交 | 1 两线段严格相交
    int is_inter(const segment<T> &s) const
    {
        if (is_on(s.a) || is_on(s.b) || s.is_on(a) || s.is_on(b))
            return -1;
        const line<T> l{a, b - a}, ls{s.a, s.b - s.a};
        return l.toleft(s.a) * l.toleft(s.b) == -1 && ls.toleft(a) * ls.toleft(b) == -1;
    }

    // 点到线段距离
    long double dis(const point<T> &p) const
    {
        if ((p - a) * (b - a) < -eps || (p - b) * (a - b) < -eps)
            return min(p.dis(a), p.dis(b));
        const line<T> l{a, b - a};
        return l.dis(p);
    }

    // 两线段间距离
    long double dis(const segment<T> &s) const
    {
        if (is_inter(s))
            return 0;
        return min({dis(s.a), dis(s.b), s.dis(a), s.dis(b)});
    }
};

using Segment = segment<point_t>;

// 多边形
template <typename T>
struct polygon
{
    vector<point<T>> p; // 以逆时针顺序存储

    size_t nxt(const size_t i) const { return i == p.size() - 1 ? 0 : i + 1; }
    size_t pre(const size_t i) const { return i == 0 ? p.size() - 1 : i - 1; }

    // 回转数
    // 返回值第一项表示点是否在多边形边上
    // 对于狭义多边形，回转数为 0 表示点在多边形外，否则点在多边形内
    pair<bool, int> winding(const point<T> &a) const
    {
        int cnt = 0;
        for (size_t i = 0; i < p.size(); i++)
        {
            const point<T> u = p[i], v = p[nxt(i)];
            if (abs((a - u) ^ (a - v)) <= eps && (a - u) * (a - v) <= eps)
                return {true, 0};
            if (abs(u.y - v.y) <= eps)
                continue;
            const Line uv = {u, v - u};
            if (u.y < v.y - eps && uv.toleft(a) <= 0)
                continue;
            if (u.y > v.y + eps && uv.toleft(a) >= 0)
                continue;
            if (u.y < a.y - eps && v.y >= a.y - eps)
                cnt++;
            if (u.y >= a.y - eps && v.y < a.y - eps)
                cnt--;
        }
        return {false, cnt};
    }

    // 多边形面积的两倍
    // 可用于判断点的存储顺序是顺时针或逆时针
    T area() const
    {
        T sum = 0;
        for (size_t i = 0; i < p.size(); i++)
            sum += p[i] ^ p[nxt(i)];
        return abs(sum);
    }

    // 多边形的周长
    long double circ() const
    {
        long double sum = 0;
        for (size_t i = 0; i < p.size(); i++)
            sum += p[i].dis(p[nxt(i)]);
        return sum;
    }
};

using Polygon = polygon<point_t>;

// 凸多边形
template <typename T>
struct convex : polygon<T>
{
    // 闵可夫斯基和
    convex operator+(const convex &c) const
    {
        const auto &p = this->p;
        vector<Segment> e1(p.size()), e2(c.p.size()), edge(p.size() + c.p.size());
        vector<point<T>> res;
        res.reserve(p.size() + c.p.size());
        const auto cmp = [](const Segment &u, const Segment &v)
        { return argcmp()(u.b - u.a, v.b - v.a); };
        for (size_t i = 0; i < p.size(); i++)
            e1[i] = {p[i], p[this->nxt(i)]};
        for (size_t i = 0; i < c.p.size(); i++)
            e2[i] = {c.p[i], c.p[c.nxt(i)]};
        rotate(e1.begin(), min_element(e1.begin(), e1.end(), cmp), e1.end());
        rotate(e2.begin(), min_element(e2.begin(), e2.end(), cmp), e2.end());
        merge(e1.begin(), e1.end(), e2.begin(), e2.end(), edge.begin(), cmp);
        const auto check = [](const vector<point<T>> &res, const point<T> &u)
        {
            const auto back1 = res.back(), back2 = *prev(res.end(), 2);
            return (back1 - back2).toleft(u - back1) == 0 && (back1 - back2) * (u - back1) >= -eps;
        };
        auto u = e1[0].a + e2[0].a;
        for (const auto &v : edge)
        {
            while (res.size() > 1 && check(res, u))
                res.pop_back();
            res.push_back(u);
            u = u + v.b - v.a;
        }
        if (res.size() > 1 && check(res, res[0]))
            res.pop_back();
        return {res};
    }

    // 旋转卡壳
    // func 为更新答案的函数，可以根据题目调整位置
    template <typename F>
    void rotcaliper(const F &func) const
    {
        const auto &p = this->p;
        const auto area = [](const point<T> &u, const point<T> &v, const point<T> &w)
        { return (w - u) ^ (w - v); };
        for (size_t i = 0, j = 1; i < p.size(); i++)
        {
            const auto nxti = this->nxt(i);
            func(p[i], p[nxti], p[j]);
            while (area(p[this->nxt(j)], p[i], p[nxti]) >= area(p[j], p[i], p[nxti]))
            {
                j = this->nxt(j);
                func(p[i], p[nxti], p[j]);
            }
        }
    }

    // 凸多边形的直径的平方
    T diameter2() const
    {
        const auto &p = this->p;
        if (p.size() == 1)
            return 0;
        if (p.size() == 2)
            return p[0].dis2(p[1]);
        T ans = 0;
        auto func = [&](const point<T> &u, const point<T> &v, const point<T> &w)
        { ans = max({ans, w.dis2(u), w.dis2(v)}); };
        rotcaliper(func);
        return ans;
    }

    // 判断点是否在凸多边形内
    // 复杂度 O(logn)
    // -1 点在多边形边上 | 0 点在多边形外 | 1 点在多边形内
    int is_in(const point<T> &a) const
    {
        const auto &p = this->p;
        if (p.size() == 1)
            return a == p[0] ? -1 : 0;
        if (p.size() == 2)
            return segment<T>{p[0], p[1]}.is_on(a) ? -1 : 0;
        if (a == p[0])
            return -1;
        if ((p[1] - p[0]).toleft(a - p[0]) == -1 || (p.back() - p[0]).toleft(a - p[0]) == 1)
            return 0;
        const auto cmp = [&](const Point &u, const Point &v)
        { return (u - p[0]).toleft(v - p[0]) == 1; };
        const size_t i = lower_bound(p.begin() + 1, p.end(), a, cmp) - p.begin();
        if (i == 1)
            return segment<T>{p[0], p[i]}.is_on(a) ? -1 : 0;
        if (i == p.size() - 1 && segment<T>{p[0], p[i]}.is_on(a))
            return -1;
        if (segment<T>{p[i - 1], p[i]}.is_on(a))
            return -1;
        return (p[i] - p[i - 1]).toleft(a - p[i - 1]) > 0;
    }

    // 凸多边形关于某一方向的极点
    // 复杂度 O(logn)
    // 参考资料：https://codeforces.com/blog/entry/48868
    template <typename F>
    size_t extreme(const F &dir) const
    {
        const auto &p = this->p;
        const auto check = [&](const size_t i)
        { return dir(p[i]).toleft(p[this->nxt(i)] - p[i]) >= 0; };
        const auto dir0 = dir(p[0]);
        const auto check0 = check(0);
        if (!check0 && check(p.size() - 1))
            return 0;
        const auto cmp = [&](const Point &v)
        {
            const size_t vi = &v - p.data();
            if (vi == 0)
                return 1;
            const auto checkv = check(vi);
            const auto t = dir0.toleft(v - p[0]);
            if (vi == 1 && checkv == check0 && t == 0)
                return 1;
            return checkv ^ (checkv == check0 && t <= 0);
        };
        return partition_point(p.begin(), p.end(), cmp) - p.begin();
    }

    // 过凸多边形外一点求凸多边形的切线，返回切点下标
    // 复杂度 O(logn)
    // 必须保证点在多边形外
    pair<size_t, size_t> tangent(const point<T> &a) const
    {
        const size_t i = extreme([&](const point<T> &u)
                                 { return u - a; });
        const size_t j = extreme([&](const point<T> &u)
                                 { return a - u; });
        return {i, j};
    }

    // 求平行于给定直线的凸多边形的切线，返回切点下标
    // 复杂度 O(logn)
    pair<size_t, size_t> tangent(const line<T> &a) const
    {
        const size_t i = extreme([&](...)
                                 { return a.v; });
        const size_t j = extreme([&](...)
                                 { return -a.v; });
        return {i, j};
    }
};

using Convex = convex<point_t>;

// 点集的凸包
// Andrew 算法，复杂度 O(nlogn)
Convex convexhull(vector<Point> p)
{
    vector<Point> st;
    if (p.empty())
        return Convex{st};
    if (p.size() == 1)
        return Convex{p};
    sort(p.begin(), p.end());
    const auto check = [](const vector<Point> &st, const Point &u)
    {
        const auto back1 = st.back(), back2 = *prev(st.end(), 2);
        return (back1 - back2).toleft(u - back1) <= 0;
    };
    for (const Point &u : p)
    {
        while (st.size() > 1 && check(st, u))
            st.pop_back();
        st.push_back(u);
    }
    size_t k = st.size();
    p.pop_back();
    reverse(p.begin(), p.end());
    for (const Point &u : p)
    {
        while (st.size() > k && check(st, u))
            st.pop_back();
        st.push_back(u);
    }
    st.pop_back();
    return Convex{st};
}

bool _output = 1;

void solve()
{
    int n;
    cin >> n;
    vector<Point> a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].x >> a[i].y;
    }
    sort(a.begin(), a.end());
    auto p1 = convexhull(a);
    auto c1 = p1.p;
    if (c1.size() == n)
    {
        cout << -1 << endl;
        return;
    }
    set<Point> st;
    for (int i = 0; i < c1.size(); i++)
    {
        st.insert(c1[i]);
    }
    // cout << "! ";
    // for (auto x : st)
    // {
    //     cout << x.x << " " << x.y << endl;
    // }
    int st_area = p1.area();
    vector<Point> b;
    for (int i = 0; i < n; i++)
    {
        if (!st.count(a[i]))
            b.push_back(a[i]);
    }
    auto p2 = convexhull(b);
    auto c2 = p2.p;
    // cout << c2.size() << endl;
    int sz = c1.size();
    int pos = 0;
    auto make = [&](Point p1, Point p2, Point p3)
    {
        return (ll)abs((p2 - p1) ^ (p3 - p1));
    };
    ll ans = 0;

    for (int i = 0; i < c1.size(); i++)
    {
        auto p1 = c1[i];
        auto p2 = c1[(i + 1) % sz];
        auto p3 = c2[pos];
        auto p4 = c2[(pos + 1) % c2.size()];
        while (make(p1, p2, p3) > make(p1, p2, p4))
        {
            pos = (pos + 1) % c2.size();
            p3 = c2[pos];
            p4 = c2[(pos + 1) % c2.size()];
        }
        ans = max(ans, st_area - make(p1, p2, p3));
    }
    cout << ans << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int _ = 1;
    if (_output)
        cin >> _;
    while (_--)
        solve();
    return 0;
}