#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
using namespace std;
int gcd(int a, int b) {
  if (!a || !b) return max(a, b);
  return gcd(max(a, b) % min(a, b), min(a, b));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (int k = 2; k <= 1000; k++) {
      int a = n - (k + 1);
      int b = k;
      int c = 1;
      if (a < 0 || b < 0 || c < 0) continue;
      if (a == b || b == c || a == c) continue;
      if (gcd(a, b) != c) continue;
      cout << a << " " << b << " " << c << '\n';
      break;
    }
  }
}

### Prompt

Create a solution in CPP for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
using namespace std;
int gcd(int a, int b) {
  if (!a || !b) return max(a, b);
  return gcd(max(a, b) % min(a, b), min(a, b));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (int k = 2; k <= 1000; k++) {
      int a = n - (k + 1);
      int b = k;
      int c = 1;
      if (a < 0 || b < 0 || c < 0) continue;
      if (a == b || b == c || a == c) continue;
      if (gcd(a, b) != c) continue;
      cout << a << " " << b << " " << c << '\n';
      break;
    }
  }
}

```