// L素数が本質なのはすぐ分かるが、そこからが苦手。
// このテーブルを眺めていたら、aが素数のとき、a=m+n, m=n+1が見えた。
// https://mathlandscape.com/pythagoras-triple-table/
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <tuple>
#include <cstdio>
#include <cassert>
//#include <atcoder/modint>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;
//using namespace atcoder;

// https://algo-method.com/tasks/553/editorial
// Miller-Rabin 素数判定法
template<class T> T pow_mod(T A, T N, T M) {
    T res = 1 % M;
    A %= M;
    while (N) {
        if (N & 1) res = (res * A) % M;
        A = (A * A) % M;
        N >>= 1;
    }
    return res;
}

bool is_prime(long long N) {
    if (N <= 1) return false;
    if (N == 2 || N == 3) return true;
    if (N % 2 == 0) return false;
    vector<long long> A = {2, 325, 9375, 28178, 450775,
                           9780504, 1795265022};
    long long s = 0, d = N - 1;
    while (d % 2 == 0) {
        ++s;
        d >>= 1;
    }
    for (auto a : A) {
        if (a % N == 0) return true;
        long long t, x = pow_mod<__int128_t>(a, d, N);
        if (x != 1) {
            for (t = 0; t < s; ++t) {
                if (x == N - 1) break;
                x = __int128_t(x) * x % N;
            }
            if (t == s) return false;
        }
    }
    return true;
}

// Pollard のロー法
long long gcd(long long A, long long B) {
    A = abs(A), B = abs(B);
    if (B == 0) return A;
    else return gcd(B, A % B);
}
    
long long pollard(long long N) {
    if (N % 2 == 0) return 2;
    if (is_prime(N)) return N;

    long long step = 0;
    auto f = [&](long long x) -> long long {
        return (__int128_t(x) * x + step) % N;
    };
    while (true) {
        ++step;
        long long x = step, y = f(x);
        while (true) {
            long long p = gcd(y - x + N, N);
            if (p == 0 || p == N) break;
            if (p != 1) return p;
            x = f(x);
            y = f(f(y));
        }
    }
}

vector<long long> prime_factorize(long long N) {
    if (N == 1) return {};
    long long p = pollard(N);
    if (p == N) return {p};
    vector<long long> left = prime_factorize(p);
    vector<long long> right = prime_factorize(N / p);
    left.insert(left.end(), right.begin(), right.end());
    sort(left.begin(), left.end());
    return left;
}


void solve(int L) {
    if (L % 4 == 0) {
        cout << (L / 4) * 3 << " " << (L / 4) * 4 << " " << (L / 4) * 5 << endl;
        return;
    }

    vector<int> ps = prime_factorize(L);
    int p = ps.back();
    int m = (p + 1) / 2, n = p / 2;
    int bai = L / p;
    cout << (m * m - n * n) * bai << " " << (m * m + n * n) * bai << " " << 2 * m * n * bai << endl;
}

signed main() {
    int T;
    cin >> T;
    while (T--) {
        int L; cin >> L;
        solve(L);
    }
    return 0;
}