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 inf = std::numeric_limits<long long>::max();
  void setIO(string name = "") {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    if (name != "") {
      freopen((name + ".in").c_str(), "r", stdin);
      freopen((name + ".out").c_str(), "w", stdout);
    }
  }
  int main() {
    setIO("");
    long long T;
    cin >> T;
    cin.ignore();
    while (T--) {
      long long n;
      cin >> n;
      if (n & 1) {
        long long k = (n - 1) / 2;
        long long r = n / 2;
        if (k & 1) {
          cout << r - 2 << " " << r + 2 << " " << 1 << "\n";
        } else {
          cout << r - 1 << " " << r + 1 << " " << 1 << "\n";
        }
      } else {
        cout << 2 << " " << n - 3 << " " << 1 << "\n";
      }
    }
    return 0;
  }
  