description: Given a positive integer n. Find three distinct positive integers a, b, c such that a + b + c = n and \operatorname{gcd}(a, b) = c, where \operatorname{gcd}(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
  
  The first and only line of each test case contains a single integer n (10 ≤ n ≤ 10^9).
  
  Output
  
  For each test case, output three distinct positive integers a, b, c satisfying the requirements. If there are multiple solutions, you can print any. We can show that an answer always exists.
  
  Example
  
  Input
  
  
  6
  18
  63
  73
  91
  438
  122690412
  
  
  Output
  
  
  6 9 3
  21 39 3
  29 43 1
  49 35 7
  146 219 73
  28622 122661788 2
  
  Note
  
  In the first test case, 6 + 9 + 3 = 18 and \operatorname{gcd}(6, 9) = 3.
  
  In the second test case, 21 + 39 + 3 = 63 and \operatorname{gcd}(21, 39) = 3.
  
  In the third test case, 29 + 43 + 1 = 73 and \operatorname{gcd}(29, 43) = 1.
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long K = 998244353, N = 6 * 1e5 + 1, INF = 2e17;
  vector<long long> d;
  vector<vector<long long>> g;
  vector<long long> w;
  bool ok = 1;
  long long bp(long long n, long long p) {
    if (p == 0) return 1;
    if (p % 2 == 0) {
      long long res = bp(n, p / 2) % K;
      return (res * res) % K;
    } else {
      return (n * (bp(n, p - 1)) % K) % K;
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      if (n % 2 == 0) {
        cout << n / 2 - 1 << " " << n / 2 << " " << 1 << "\n";
        continue;
      }
      if (n % 3 == 2) {
        cout << (n - 2) / 3 << " " << (2 * (n - 2) / 3 + 1) << " 1\n";
        continue;
      }
      if (n % 3 == 0) {
        long long k = 0;
        if ((n - 3) % 2 == 0) k = 1;
        cout << (n - 3) / 3 + k << " " << 2 * (n - 3) / 3 + 2 - k << " 1\n";
        continue;
      }
      if (n % 3 == 1) {
        long long k = 0;
        if ((n + 1) % 4 == 0) {
          k = 1;
        }
        cout << (n + 1) / 2 - 2 - k << " " << (n + 1) / 2 + k << " 1\n";
        continue;
      }
    }
  }
  