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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  template <typename A, typename B>
  A min(A x, B y) {
    if (x < y) return x;
    return y;
  }
  template <typename A, typename B>
  A max(A x, B y) {
    if (x > y) return x;
    return y;
  }
  template <typename D>
  istream &operator>>(istream &in, vector<D> &arr) {
    for (auto &(x) : (arr)) cin >> x;
    return in;
  }
  template <typename D>
  ostream &operator<<(ostream &in, vector<D> &arr) {
    for (auto &x : arr) in << x << " ";
    return in;
  }
  template <typename D, typename B>
  ostream &operator<<(ostream &in, pair<D, B> &p) {
    in << p.first << " " << p.second << " ";
    return in;
  }
  template <typename D, typename B>
  istream &operator>>(istream &in, pair<D, B> &p) {
    in >> p.first >> p.second;
    return in;
  }
  int modpow(long long, long long, int);
  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 << "]";
  }
  void solve(int tc) {
    int n;
    cin >> n;
    if (n % 2 == 0) {
      cout << (n + 1) / 2 << " " << (n - 1) / 2 << " " << 1 << " "
           << "\n";
      return;
    }
    n /= 2;
    if (n % 2 == 0) {
      cout << n - 1 << " " << n + 1 << " " << 1 << " "
           << "\n";
    } else {
      cout << n - 2 << " " << n + 2 << " " << 1 << " "
           << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    int k = 1, t = 1;
    cin >> t;
    while (t--) solve(k++);
    return 0;
  }
  int modpow(long long a, long long b, int m = 1000000007) {
    long long res = 1;
    a %= m;
    if (a == 0) return a;
    while (b > 0) {
      if (b & 1) res = (res * a) % m;
      a *= a;
      a %= m;
      b >>= 1;
    }
    return res;
  }
  