#include<bits/stdc++.h>
#define int long long
#define double long double
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<int,int> PII;
#define db double
const db eps = 1e-12;
const db inf = 1e20;
const db PI = acos(-1.0);
inline int sgn(db x) {return x < -eps ? -1 : x > eps;}
inline db sqr(db x) { return x*x; }
inline db mysqrt(db x) { return sqrt(max((double)0.0, x)); }
//注意y1给系统占了
const int N=3e5+10;
const int M=1e3+10;
int mod=1e9+7;
/*========================================================*\
| Point 类
| 							类内函数
| Point(P x = 0, P y = 0)					- 默认构造函数
| input()									- 读入一个点
| operator == 								- NULL
| operator <								- 用于排序
| operator +								- 返回Point
| operator -								- 返回Point
| operator *								- 返回Point
| operator /								- 返回Point
| len()										- 返回长度
| len2()									- 返回长度平方		
| trunc(db len)								- 长度变为len
| rotleft()									- 逆时针转π/2, 返回Point
| rotright()								- 顺时针转π/2, 返回Point
| rotate(P p, db ang)						- 绕p逆时针转ang
| 							类外函数
| *基础功能
| Cross(P a, P b)							- NULL
| Cross(P pt, P a, P b)						- NULL
| Dot(P a, P b)								- NULL
| Dot(P pt, P a, P b)						- NULL
| Dist(P a, P b)							- NULL
| Len(a)									- 返回a的长度
| Len2(a) 									- 返回a的长度平方
| rad(P p, P a, P b) 						- 返回∠apb	
| Complex类补充
\*========================================================*/
struct Point {
    db x, y;
    Point (db x=0, db y=0): x(x), y(y) {}
    void input() { scanf("%lf%lf", &x, &y); }
    bool operator == (Point r) { return sgn(x-r.x)==0 && sgn(y-r.y)==0; }
    // bool operator < (Point r) { return sgn(x-r.x)==0 ? y < r.y : x < r.x; }
    Point operator + (Point r) { return Point (x+r.x, y+r.y); }
    Point operator - (Point r) { return Point (x-r.x, y-r.y); }
    Point operator * (db r) { return Point (x*r, y*r); }
    Point operator / (db r) { return Point (x/r, y/r); }
    db len() { return mysqrt(x*x+y*y); }
    db len2() { return x*x + y*y; }
    // 化为长度为 r 的向量
    Point trunc(db len) {
        db l = mysqrt(x * x + y * y);
        if (! sgn(l)) return *this;
        len /= l;
        return Point (x*len, y*len);
    }
    // 逆时针旋转 90 度
    Point rotleft() { return Point (-y, x); }
    // 顺时针旋转 90 度
    Point rotright() { return Point (y, -x); }
    // 绕着 p 点逆时针旋转 ang
    Point rotate (Point p, db ang) {
        Point v = (*this) - p;
        db c = cos(ang), s = sin(ang);
        return Point (p.x + v.x*c - v.y*s, p.y + v.x*s + v.y*c);
    }
};
/******************************\
| 			基础功能
\******************************/
db Cross (Point a, Point b) { return a.x*b.y - a.y*b.x; }
db Cross (Point pt, Point a, Point b) { return Cross(a-pt, b-pt); }
db Dot (Point a, Point b) { return a.x*b.x + a.y*b.y; }
db Dot (Point pt, Point a, Point b) { return Dot (a-pt, b-pt); }
db Dist (Point a, Point b) { return mysqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }
db Len (Point a) { return mysqrt(a.x*a.x + a.y*a.y); }
db Len2 (Point a) { return a.x*a.x + a.y*a.y; }
// 计算 pa 和 pb 的夹角,  就是求这个点看 a, b 所成的夹角, 测试 LightOJ 1203
db rad(Point p, Point a, Point b) { 
    return fabs(atan2(fabs(Cross(p, a, b)), Dot(p, a, b) )); 
}
/******************************\
| 			Complex补充
\******************************/
struct Complex {
    db x, y;
    Complex(db x = 0, db y = 0) : x(x), y(y) {}
    Complex operator + (Complex b) {return Complex(x + b.x, y + b.y);}
    Complex operator - (Complex b) {return Complex(x - b.x, y - b.y);}
    Complex operator * (Complex b) {return Complex(x * b.x - y * b.y, x * b.y + y * b.x);}
};
// db disptl(Line v, Point p) {
//     return fabs(Cross(v.s, p, v.e)) / v.length(); // length = [](){return Dist(v.s, v.e);}
// }
// Andrew 算法
// 坐标排序 O(nlogn), 测试: 洛谷P2742；凸包稳定相关: POJ1228
// 这里有两种排序方法
// 1. 优先按 y 排序, 如果 y 相同, 按 x 排序, 起点是最下面的点, 如下面代码体现
// 2. 优先按 x 排序, 如果 x 相同, 按 y 排序, 起点是最左边的点
bool operator < (Point a, Point b) { 
	if (sgn(a.y - b.y) == 0) return a.x < b.x;
    return a.y < b.y;
}
int andrew(Point *p, Point *ch, int n) {  // 返回值是凸包的顶点数
    sort (p, p + n);   // 排序
    n = unique(p, p + n) - p; // 去除重复的点
    int v = 0;
    // 求下凸包, 如果 p[i] 是右拐弯的, 则这个点不在凸包中, 往回退
    for (int i = 0; i < n; i ++) {
        while (v > 1 && sgn(Cross(ch[v-2], ch[v-1], p[i])) <= 0) v --;
        ch[v ++] = p[i];
    }
    int j = v;
    // 求上凸包
    for (int i = n - 2; i >= 0; i --) {
        while (v > j && sgn(Cross(ch[v-2], ch[v-1], p[i])) <= 0) v --;
        ch[v ++] = p[i];
    }
    if (n > 1) v --;
    return v;
}
db area(Point a, Point b, Point c){
    Point AB=b-a,AC=c-a;
    return fabs(AB.x*AC.y-AB.y*AC.x);
}
db polygonarea(Point* p, int n) {
    db ans = 0;
    p[n] = p[0];
    for (int i = 0; i < n; i ++) 
        ans += Cross(p[i], p[i+1]);
    return fabs(ans) ;           // 面积有正负, 需根据题意来判断要不要取绝对值
}
db disptl(Point A, Point B, Point p) {
    return fabs(Cross(A, p, B)) / Dist(A, B); // length = [](){return Dist(v.s, v.e);}
}
Point ch1[N], ch2[N],p1[N],p2[N];
int vis[N];
void solve(){
    int n;cin>>n;
    for(int i=0;i<=n;i++) p1[i].x=p1[i].y=p2[i].x=p2[i].y=ch1[i].x=ch1[i].y=ch2[i].x=ch2[i].y=vis[i]=0;
    map<PII,int>mp;
    for(int i=0;i<n;i++) cin>>p1[i].x>>p1[i].y;
    int m1 = andrew(p1, ch1, n);
    if(m1==n){
        cout<<"-1\n";
        return ;
    }
    for(int i=0;i<n;i++) mp[{p1[i].x,p1[i].y}]=i;
    // cout<<"\n";
    for(int i=0;i<m1;i++){
        vis[mp[{ch1[i].x,ch1[i].y}]]=1;
    }
    int sp=0;
    for(int i=0;i<n;i++){
        if(!vis[i]) p2[sp++]=p1[i];
        // cout<<vis[i]<<" ";
    }
    int m2 = andrew(p2, ch2, sp);
    double S=polygonarea(ch1, m1);
    int idx=0;
    double ans=0;
    // cout<<"\n";
    for(int i=0;i<m1;i++){
        double fz=1e19,mu=1;
        Point A=ch1[i],B=ch1[(i+1)%m1];
        while(1){
            double fz1=fabs(Cross(A,B,ch2[idx]))*fabs(Cross(A,B,ch2[idx]));
            double mu1=(A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);
            if(fz1*mu<fz*mu1) fz=fz1,mu=mu1;
            else break;
            idx=(idx+1)%m2;
        }
        idx=(idx-1+m2)%m2;
        ans=max(ans,S-area(A,B,ch2[idx]));
    }
    cout<<(int)ans<<"\n";
}

signed main(){
    // ios::sync_with_stdio(false);
    // cin.tie(nullptr),cout.tie(nullptr);
    int _;
    _=1;
    cin>>_;
    while(_--){
        solve();
    }
}