#include <bits/stdc++.h>
using namespace std;
int gys(int x, int y) {
  if (!y)
    return x;
  else
    return gys(y, x % y);
}
int main() {
  int t, k;
  int a, b, c = 1;
  cin >> t;
  while (t--) {
    cin >> k;
    k--;
    for (b = 2; 2 * b < k; ++b) {
      if (gys(k - b, b) == 1) break;
    }
    a = k - b;
    printf("%d %d %d\n", a, b, c);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
using namespace std;
int gys(int x, int y) {
  if (!y)
    return x;
  else
    return gys(y, x % y);
}
int main() {
  int t, k;
  int a, b, c = 1;
  cin >> t;
  while (t--) {
    cin >> k;
    k--;
    for (b = 2; 2 * b < k; ++b) {
      if (gys(k - b, b) == 1) break;
    }
    a = k - b;
    printf("%d %d %d\n", a, b, c);
  }
  return 0;
}

```