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;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  int mod_add(int a, int b, int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  int mod_mul(int a, int b, int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  int mod_sub(int a, int b, int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  const long long mxm = 2 * 1e5;
  long long M = 1e9 + 7;
  long long gcd(long long a, long long b) {
    if (b == 0) {
      return a;
    } else {
      return gcd(b, a % b);
    }
  }
  long long expo(long long a, long long b, long long m) {
    if (b == 0) {
      return 1;
    }
    if (b % 2 == 0) {
      long long temp = expo(a, b / 2, m);
      return (temp % m * temp % m) % m;
    } else {
      long long temp = expo(a, b / 2, m);
      return (a % m * temp % m * temp % m) % m;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    t = 1;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      if (n % 2 == 0) {
        cout << n / 2 << ' ' << n / 2 - 1 << ' ' << 1 << "\n";
      } else {
        long long a = (n + 1) / 2;
        long long b = n / 2;
        if (b % 2 == 0) {
          cout << (n + 1) / 2 << ' ' << (n / 2) - 1 << ' ' << 1 << ' ' << "\n";
        } else {
          cout << (n + 1) / 2 + 1 << ' ' << (n / 2) - 2 << ' ' << 1 << "\n";
        }
      }
    }
    return 0;
  }
  