#include <bits/stdc++.h>
using namespace std;
long long mod = pow(10, 9) + 7;
long long power(long long n, long long k) {
  long long res = 1;
  while (k) {
    if (k % 2 != 0) {
      res *= n;
    }
    n = n * n;
    k /= 2;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long x = n - 1;
  long long y = x / 2;
  if (x % 2 == 1) {
    cout << y + 1 << " " << y << " 1";
  } else {
    if (y % 2 == 1) {
      cout << y + 2 << " " << y - 2 << " 1";
    } else
      cout << y + 1 << " " << y - 1 << " 1";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}

### 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;
long long mod = pow(10, 9) + 7;
long long power(long long n, long long k) {
  long long res = 1;
  while (k) {
    if (k % 2 != 0) {
      res *= n;
    }
    n = n * n;
    k /= 2;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long x = n - 1;
  long long y = x / 2;
  if (x % 2 == 1) {
    cout << y + 1 << " " << y << " 1";
  } else {
    if (y % 2 == 1) {
      cout << y + 2 << " " << y - 2 << " 1";
    } else
      cout << y + 1 << " " << y - 1 << " 1";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}

```