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;
  vector<int> vi[200005];
  vector<int>::iterator child;
  vector<int> vi2[200005];
  vector<int> vi3[200005];
  vector<long long int> vl;
  vector<vector<pair<int, int> > > vii;
  vector<pair<long long int, long long int> > vll;
  int arr[200005];
  int arr2[200005];
  bool check[200005];
  bool check2[1000][1000];
  int n, m;
  int x[] = {1, -1, 0, 0};
  int y[] = {0, 0, -1, 1};
  int x2[] = {1, -1, 0, 0, 1, -1, 1, -1};
  int y2[] = {0, 0, -1, 1, 1, -1, -1, 1};
  int bx[] = {0, 0, 1, -1, -1, 1, 1, -1};
  int by[] = {1, -1, 0, 0, -1, -1, 1, 1};
  int kx[] = {1, 1, 2, 2, -1, -1, -2, -2};
  int ky[] = {2, -2, 1, -1, 2, -2, 1, -1};
  bool isvisited(int xx, int yy) {
    if (xx < 1 || yy < 1 || xx > n || yy > m) {
      return false;
    } else if (check2[xx][yy] == true) {
      return false;
    } else
      return true;
  }
  int gcd(int a, int b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  int lcm(int a, int b) {
    int res = (a * b) / gcd(a, b);
    return res;
  }
  int po(int a, int b) {
    int res = 1;
    while (b) {
      if (b % 2 != 0) {
        res *= a;
        b--;
      } else {
        a *= a;
        b /= 2;
      }
    }
    return res;
  }
  long long int modMul(long long int a, long long int b, long long int mod) {
    long long int ans = 0;
    a = a % mod;
    while (b > 0) {
      if (b % 2) ans = (ans % mod + a % mod) % mod;
      a = (a % mod * 2 % mod) % mod;
      b /= 2;
    }
    return ans % mod;
  }
  long long int powerMod(long long int a, long long int b, long long int mod) {
    if (b == 0) return 1LL % mod;
    long long int x = powerMod(a, b / 2, mod);
    x = (x % mod * x % mod) % mod;
    if (b % 2 == 1) x = (x % mod * a % mod) % mod;
    return x % mod;
  }
  int main() {
    int test = 1;
    scanf("%d", &test);
    for (int xx = 1; xx <= test; xx++) {
      scanf("%d", &n);
      if (n % 2 == 0) {
        cout << n - 3 << " " << 2 << " " << 1 << endl;
      } else {
        n /= 2;
        if (n % 2 == 1) {
          cout << n + 2 << " " << n - 2 << " " << 1 << endl;
        } else {
          cout << n + 1 << " " << n - 1 << " " << 1 << endl;
        }
      }
    }
  }
  