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 MOD = 1e9 + 7;
  const long long MOD2 = 998244353;
  long long power(long long a, long long b) {
    long long res = 1;
    a = a % MOD;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % MOD;
      }
      a = (a * a) % MOD;
      b >>= 1;
    }
    return res;
  }
  long long fermat_inv(long long y) { return power(y, MOD - 2); }
  long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
  void solveQuestion() {
    long long n;
    cin >> n;
    if (n % 2 == 0) {
      long long a = (n - 1) / 2, b = a + 1, c = 1;
      cout << a << " " << b << " " << c << "\n";
    } else {
      if (n % 3 != 1) {
        cout << "3 " << n - 4 << " 1\n";
      } else {
        long long div = (n - 1) / 3;
        if (div % 4 == 0) {
          long long a = (n - 2) / 2, b = a + 2, c = 1;
          cout << a << " " << b << " " << c << "\n";
        } else {
          long long a = (n - 1) / 2 - 2, b = a + 4, c = 1;
          cout << a << " " << b << " " << c << "\n";
        }
      }
    }
  }
  int32_t main(int32_t argc, char const *argv[]) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long tt = 1;
    cin >> tt;
    while (tt-- > 0) {
      solveQuestion();
    }
    return 0;
  }
  