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 M = 1000000007;
  long long gcd(long long a, long long b);
  long long fact(long long n);
  long long nck(long long n, long long k);
  long long binS(vector<long long> arr, long long l, long long r, long long x);
  long long mult(long long a, long long b);
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    long long kl;
    cin >> kl;
    while (kl--) {
      long long n;
      cin >> n;
      long long a, b, c;
      if (n % 2 == 0) {
        c = 1;
        n--;
        a = n / 2;
        b = n / 2 + 1;
      } else {
        if (((n - 3) / 2) % 2 != 0) {
          c = 1;
          b = (n - 3) / 2;
          a = b + 2;
        } else {
          c = 1;
          b = (n - 3) / 2 - 1;
          a = b + 4;
        }
      }
      if (gcd(a, b) == c) {
        cout << a << " " << b << " " << c << endl;
      }
    }
    return 0;
  }
  long long binS(vector<long long> arr, long long l, long long r, long long x) {
    if (r >= l) {
      long long mid = l + (r - l) / 2;
      if (arr[mid] == x) return mid;
      if (arr[mid] > x) return binS(arr, l, mid - 1, x);
      return binS(arr, mid + 1, r, x);
    }
    return -1;
  }
  long long nck(long long n, long long k) {
    long long ans = 1;
    for (long long i = n; i > max(k, n - k); i--) {
      ans *= i;
    }
    for (long long i = 2; i <= min(k, n - k); i++) ans /= i;
    return ans;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long fact(long long n) {
    long long ans = 1;
    for (long long i = 2; i <= n; i++) ans *= i;
    return ans;
  }
  long long mult(long long a, long long b) {
    long long ans = 1;
    while (b--) ans *= a;
    return ans;
  }
  