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 MINll = 1e9;
  struct stct {};
  void sieve_of_prime(int n, vector<int>* prime) {
    bool mark[n + 1];
    memset(mark, true, sizeof(mark));
    mark[0] = mark[1] = false;
    for (int i = 2; i * i <= n; i++)
      if (mark[i])
        for (int j = 2 * i; j <= n; j += i) mark[j] = false;
    for (int i = 2; i <= n; i++)
      if (mark[i]) prime->push_back(i);
  }
  unsigned long long P = 1000000000 + 7;
  long long MAX = -1000000000 - 1;
  vector<int> prime;
  unsigned long long power(unsigned long long a, unsigned long long b) {
    unsigned long long res = 1;
    a = a % P;
    if (a == 0) return 0;
    while (b) {
      if (b % 2 == 1) res = (res * a) % P;
      a = (a * a) % P;
      b = b >> 1;
    }
    return res;
  }
  unsigned long long bsearch(unsigned long long l, unsigned long long r,
                             unsigned long long x, unsigned long long k) {
    if (l >= r) return l;
    unsigned long long m = (l + r) / 2;
    unsigned long long sum =
        m > k ? k * k - (2 * k - m) * (2 * k - m - 1) / 2 : m * (m + 1) / 2;
    if (sum < x)
      return bsearch(m + 1, r, x, k);
    else
      return bsearch(l, m, x, k);
  }
  void solve() {
    int n;
    cin >> n;
    int a = 1, b = 1, c = 1;
    n--;
    a = b = n / 2;
    if (n % 2 == 0) {
      if (n % 4 == 0) {
        a--;
        b++;
      } else {
        a -= 2;
        b += 2;
      }
    } else
      b++;
    cout << a << " " << b << " " << c << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    sieve_of_prime(1e5, &prime);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  