#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define per(i, j, k) for (int i = (j); i >= (k); --i)
#define sz(v) int((v).size())
#define all(v) (v).begin(), (v).end()
#define endl "\n"
#define no cout << "NO" << endl;
#define yes cout << "YES" << endl;
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;
typedef double db;
const db eps = 1e-8;
#define inf 1e18
#define ret                 \
    {                       \
        cout << -1 << endl; \
        return;             \
    }
template <typename T>
bool gmax(T &x, const T y)
{
    if (x < y)
        return x = y, 1;
    return 0;
}
template <typename T>
bool gmin(T &x, const T y)
{
    if (y < x)
        return x = y, 1;
    return 0;
}
#define maxn 200003
#define int long long
void solve()
{
    int T;
    cin >> T;
    int num1 = floor(sqrt(1.0 * 2 * T)), num2 = ceil(sqrt(1.0 * 2 * T));
    int x1 = 2 * T + num1 * (num1 - 1), y1 = 2 * num1;
    int x2 = 2 * T + num2 * (num2 - 1), y2 = 2 * num2;
    int GCD;
    if (x1 * y2 < x2 * y1)
    {
        GCD = gcd(x1, y1);
        x1 /= GCD, y1 /= GCD;
        cout << x1 << " " << y1 << endl;
    }
    else
    {
        GCD = gcd(x2, y2);
        x2 /= GCD, y2 /= GCD;
        cout << x2 << " " << y2 << endl;
    }
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}
// 循环边界
// 二分l和r的范围
// 数组越界
// 在函数内开数组赋初值
// long long
// 尽量用数学语言刻画
// 从直觉到严谨分析，推导
// 冷静，冷静，冷静