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 double pi = acos(-1.0);
  const int dx[] = {0, 1, 0, -1};
  const int dy[] = {1, 0, -1, 0};
  const int limit = 5e5;
  int read() {
    int s = 0;
    char c = getchar(), lc = '+';
    while (c < '0' || '9' < c) lc = c, c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
    return lc == '-' ? -s : s;
  }
  void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + '0');
    else
      write(x / 10), putchar(x % 10 + '0');
  }
  void print(long long x, char c = '\n') {
    write(x);
    putchar(c);
  }
  const int mod = 1e9 + 7;
  const int M = 500 + 5;
  const int S = (1 << 13) + 5;
  const int N = 2e5 + 5;
  int n, m, _, k;
  long long a[N];
  long long b[N], d[N];
  signed main() {
    int T;
    while (cin >> T)
      while (T--) {
        scanf("%d", &n);
        if (n & 1) {
          int mid = n >> 1;
          if (mid & 1)
            printf("%d %d %d\n", n / 2 - 2, n / 2 + 2, 1);
          else
            printf("%d %d %d\n", n / 2 - 1, n / 2 + 1, 1);
        } else {
          printf("%d %d %d\n", n / 2, n / 2 - 1, 1);
        }
      }
    return 0;
  }
  