#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int64_t t;
  cin >> t;
  for (int64_t h = 0; h < t; h++) {
    int64_t n;
    cin >> n;
    int64_t a, b, c;
    if (n % 2 == 0) {
      a = n - 3;
      b = 2;
      c = 1;
    } else {
      c = 1;
      if ((n / 2) % 2 == 0) {
        a = (n / 2) + 1;
        b = (n / 2) - 1;
      } else {
        a = (n / 2) + 2;
        b = (n / 2) - 2;
      }
    }
    cout << a << ' ' << b << ' ' << c << '\n';
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int64_t t;
  cin >> t;
  for (int64_t h = 0; h < t; h++) {
    int64_t n;
    cin >> n;
    int64_t a, b, c;
    if (n % 2 == 0) {
      a = n - 3;
      b = 2;
      c = 1;
    } else {
      c = 1;
      if ((n / 2) % 2 == 0) {
        a = (n / 2) + 1;
        b = (n / 2) - 1;
      } else {
        a = (n / 2) + 2;
        b = (n / 2) - 2;
      }
    }
    cout << a << ' ' << b << ' ' << c << '\n';
  }
}

```