#include <bits/stdc++.h>
using namespace std;
const long long K = 998244353, N = 6 * 1e5 + 1, INF = 2e17;
vector<long long> d;
vector<vector<long long>> g;
vector<long long> w;
bool ok = 1;
long long bp(long long n, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long res = bp(n, p / 2) % K;
    return (res * res) % K;
  } else {
    return (n * (bp(n, p - 1)) % K) % K;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n % 2 == 0) {
      cout << n / 2 - 1 << " " << n / 2 << " " << 1 << "\n";
      continue;
    }
    if (n % 3 == 2) {
      cout << (n - 2) / 3 << " " << (2 * (n - 2) / 3 + 1) << " 1\n";
      continue;
    }
    if (n % 3 == 0) {
      long long k = 0;
      if ((n - 3) % 2 == 0) k = 1;
      cout << (n - 3) / 3 + k << " " << 2 * (n - 3) / 3 + 2 - k << " 1\n";
      continue;
    }
    if (n % 3 == 1) {
      long long k = 0;
      if ((n + 1) % 4 == 0) {
        k = 1;
      }
      cout << (n + 1) / 2 - 2 - k << " " << (n + 1) / 2 + k << " 1\n";
      continue;
    }
  }
}

### Prompt

Generate 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;
const long long K = 998244353, N = 6 * 1e5 + 1, INF = 2e17;
vector<long long> d;
vector<vector<long long>> g;
vector<long long> w;
bool ok = 1;
long long bp(long long n, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long res = bp(n, p / 2) % K;
    return (res * res) % K;
  } else {
    return (n * (bp(n, p - 1)) % K) % K;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n % 2 == 0) {
      cout << n / 2 - 1 << " " << n / 2 << " " << 1 << "\n";
      continue;
    }
    if (n % 3 == 2) {
      cout << (n - 2) / 3 << " " << (2 * (n - 2) / 3 + 1) << " 1\n";
      continue;
    }
    if (n % 3 == 0) {
      long long k = 0;
      if ((n - 3) % 2 == 0) k = 1;
      cout << (n - 3) / 3 + k << " " << 2 * (n - 3) / 3 + 2 - k << " 1\n";
      continue;
    }
    if (n % 3 == 1) {
      long long k = 0;
      if ((n + 1) % 4 == 0) {
        k = 1;
      }
      cout << (n + 1) / 2 - 2 - k << " " << (n + 1) / 2 + k << " 1\n";
      continue;
    }
  }
}

```