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;
  long long int bpow(long long int a, long long int b) {
    long long int ans = 1;
    while (b > 0) {
      if (b & 1) ans = (ans * a) % 1000000007;
      b /= 2;
      a = (a * a) % 1000000007;
    }
    return ans;
  }
  int main() {
    long long int testc;
    cin >> testc;
    while (testc--) {
      long long int n;
      cin >> n;
      if (n % 2 == 0) {
        cout << ((n - 1) / 2) << " " << ((n - 1) / 2) + 1 << " " << 1 << endl;
      } else if ((n / 2) % 2 == 0)
        cout << ((n - 1) / 2) - 1 << " " << ((n - 1) / 2) + 1 << " " << 1 << endl;
      else
        cout << ((n - 1) / 2) - 2 << " " << ((n - 1) / 2) + 2 << " " << 1 << endl;
    }
  }
  