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;
  bool isPrime[50000];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    vector<int> primes;
    fill(isPrime, isPrime + 50000, true);
    for (int i = 2; i < 50000; i++) {
      if (isPrime[i]) {
        primes.push_back(i);
        for (int j = i + i; j < 50000; j += i) {
          isPrime[j] = false;
        }
      }
    }
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      if (n % 2 == 0) {
        cout << n / 2 - 1 << ' ' << n / 2 << ' ' << 1 << '\n';
      } else {
        for (int p : primes) {
          if ((n - 1) % p != 0) {
            cout << p << ' ' << n - 1 - p << ' ' << 1 << '\n';
            break;
          }
        }
      }
    }
  }
  