#include <bits/stdc++.h>
using namespace std;
bool isPrime[50000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  vector<int> primes;
  fill(isPrime, isPrime + 50000, true);
  for (int i = 2; i < 50000; i++) {
    if (isPrime[i]) {
      primes.push_back(i);
      for (int j = i + i; j < 50000; j += i) {
        isPrime[j] = false;
      }
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if (n % 2 == 0) {
      cout << n / 2 - 1 << ' ' << n / 2 << ' ' << 1 << '\n';
    } else {
      for (int p : primes) {
        if ((n - 1) % p != 0) {
          cout << p << ' ' << n - 1 - p << ' ' << 1 << '\n';
          break;
        }
      }
    }
  }
}

### Prompt

Please formulate a CPP solution to 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;
bool isPrime[50000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  vector<int> primes;
  fill(isPrime, isPrime + 50000, true);
  for (int i = 2; i < 50000; i++) {
    if (isPrime[i]) {
      primes.push_back(i);
      for (int j = i + i; j < 50000; j += i) {
        isPrime[j] = false;
      }
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if (n % 2 == 0) {
      cout << n / 2 - 1 << ' ' << n / 2 << ' ' << 1 << '\n';
    } else {
      for (int p : primes) {
        if ((n - 1) % p != 0) {
          cout << p << ' ' << n - 1 - p << ' ' << 1 << '\n';
          break;
        }
      }
    }
  }
}

```