#include <bits/stdc++.h>
using namespace std;
const int max_size = 1e5 + 10;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << 2 << " " << n - 3 << " " << 1 << "\n";
    return;
  } else {
    for (int i = 2; i < n; i++) {
      if (gcd(i, (n - i - 1)) == 1) {
        cout << i << " " << (n - i - 1) << " " << 1 << "\n";
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cout << fixed << setprecision(6);
  ;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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;
const int max_size = 1e5 + 10;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << 2 << " " << n - 3 << " " << 1 << "\n";
    return;
  } else {
    for (int i = 2; i < n; i++) {
      if (gcd(i, (n - i - 1)) == 1) {
        cout << i << " " << (n - i - 1) << " " << 1 << "\n";
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cout << fixed << setprecision(6);
  ;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```