#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long MOD2 = 998244353;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
    }
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, MOD - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void solveQuestion() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    long long a = (n - 1) / 2, b = a + 1, c = 1;
    cout << a << " " << b << " " << c << "\n";
  } else {
    if (n % 3 != 1) {
      cout << "3 " << n - 4 << " 1\n";
    } else {
      long long div = (n - 1) / 3;
      if (div % 4 == 0) {
        long long a = (n - 2) / 2, b = a + 2, c = 1;
        cout << a << " " << b << " " << c << "\n";
      } else {
        long long a = (n - 1) / 2 - 2, b = a + 4, c = 1;
        cout << a << " " << b << " " << c << "\n";
      }
    }
  }
}
int32_t main(int32_t argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt = 1;
  cin >> tt;
  while (tt-- > 0) {
    solveQuestion();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 MOD = 1e9 + 7;
const long long MOD2 = 998244353;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
    }
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, MOD - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void solveQuestion() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    long long a = (n - 1) / 2, b = a + 1, c = 1;
    cout << a << " " << b << " " << c << "\n";
  } else {
    if (n % 3 != 1) {
      cout << "3 " << n - 4 << " 1\n";
    } else {
      long long div = (n - 1) / 3;
      if (div % 4 == 0) {
        long long a = (n - 2) / 2, b = a + 2, c = 1;
        cout << a << " " << b << " " << c << "\n";
      } else {
        long long a = (n - 1) / 2 - 2, b = a + 4, c = 1;
        cout << a << " " << b << " " << c << "\n";
      }
    }
  }
}
int32_t main(int32_t argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt = 1;
  cin >> tt;
  while (tt-- > 0) {
    solveQuestion();
  }
  return 0;
}

```