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;
  const int inf = 1e9 + 7;
  const double eps = 1e-6;
  const double pi = 1.00 * acos(-1.00);
  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.ff);
    cerr << ",";
    _print(p.ss);
    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 << "]";
  }
  struct custom_hash {
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return x + FIXED_RANDOM;
    }
  };
  long long binaryExponential(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans = (ans % 1000000007 * a % 1000000007) % 1000000007;
      }
      a = (a % 1000000007 * a % 1000000007) % 1000000007;
      b >>= 1;
    }
    return ans;
  }
  void seive() {
    vector<long long> prime(10000007, 1);
    prime[0] = prime[1] = 0;
    for (int i = 2; i * i <= 10000007; i++) {
      if (prime[i]) {
        for (int j = i * i; j <= 10000007; j += i) prime[j] = 0;
      }
    }
  }
  bool isprime(int n) {
    if (n < 2) return false;
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        return false;
      }
    }
    return true;
  }
  long long g, x, y;
  void extendedEuclid(long long a, long long b) {
    if (b == 0) {
      g = a;
      x = 1;
      y = 0;
    } else {
      extendedEuclid(b, a % b);
      long long temp = x;
      x = y;
      y = temp - (a / b) * y;
    }
  }
  long long ETF(long long n) {
    long long ans = n;
    for (long long i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
        ans = (ans / i) * (i - 1);
      }
      while (n % i == 0) n /= i;
    }
    if (n > 1) {
      ans = (ans / n) * (n - 1);
    }
    return ans;
  }
  long long factors(long long n) {
    long long count = 0;
    for (int i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        count++;
        if (n / i != i) {
          count++;
        }
      }
    }
    return count;
  }
  double myPow(double x, int n) {
    if (n == 0) return 1.00000;
    if (n < 0) {
      return (n % 2 == 0) ? 1.00000 / myPow(x * x, abs(n / 2))
                          : 1.000 / (x * myPow(x * x, abs(n / 2)));
    } else {
      return (n % 2 == 0) ? myPow(x * x, n / 2) : x * myPow(x * x, n / 2);
    }
  }
  bool isUnique(vector<long long> &v) {
    sort(v.begin(), v.end());
    for (long long i = 0; i < v.size(); i++) {
      if (v[i] == v[i + 1]) return false;
    }
    return true;
  }
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  void solution() {
    long long n;
    cin >> n;
    vector<int> ans{2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
    for (auto &x : ans) {
      if ((n - x - 1) % x != 0) {
        cout << n - x - 1 << ' ' << x << ' ' << 1 << "\n";
        break;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    cout << fixed << setprecision(12);
    while (t--) {
      solution();
    }
    return 0;
  }
  