#include <bits/stdc++.h>
using namespace std;
long long M = 1000000007;
long long gcd(long long a, long long b);
long long fact(long long n);
long long nck(long long n, long long k);
long long binS(vector<long long> arr, long long l, long long r, long long x);
long long mult(long long a, long long b);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long kl;
  cin >> kl;
  while (kl--) {
    long long n;
    cin >> n;
    long long a, b, c;
    if (n % 2 == 0) {
      c = 1;
      n--;
      a = n / 2;
      b = n / 2 + 1;
    } else {
      if (((n - 3) / 2) % 2 != 0) {
        c = 1;
        b = (n - 3) / 2;
        a = b + 2;
      } else {
        c = 1;
        b = (n - 3) / 2 - 1;
        a = b + 4;
      }
    }
    if (gcd(a, b) == c) {
      cout << a << " " << b << " " << c << endl;
    }
  }
  return 0;
}
long long binS(vector<long long> arr, long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binS(arr, l, mid - 1, x);
    return binS(arr, mid + 1, r, x);
  }
  return -1;
}
long long nck(long long n, long long k) {
  long long ans = 1;
  for (long long i = n; i > max(k, n - k); i--) {
    ans *= i;
  }
  for (long long i = 2; i <= min(k, n - k); i++) ans /= i;
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long fact(long long n) {
  long long ans = 1;
  for (long long i = 2; i <= n; i++) ans *= i;
  return ans;
}
long long mult(long long a, long long b) {
  long long ans = 1;
  while (b--) ans *= a;
  return ans;
}

### Prompt

Your task is to create 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 M = 1000000007;
long long gcd(long long a, long long b);
long long fact(long long n);
long long nck(long long n, long long k);
long long binS(vector<long long> arr, long long l, long long r, long long x);
long long mult(long long a, long long b);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long kl;
  cin >> kl;
  while (kl--) {
    long long n;
    cin >> n;
    long long a, b, c;
    if (n % 2 == 0) {
      c = 1;
      n--;
      a = n / 2;
      b = n / 2 + 1;
    } else {
      if (((n - 3) / 2) % 2 != 0) {
        c = 1;
        b = (n - 3) / 2;
        a = b + 2;
      } else {
        c = 1;
        b = (n - 3) / 2 - 1;
        a = b + 4;
      }
    }
    if (gcd(a, b) == c) {
      cout << a << " " << b << " " << c << endl;
    }
  }
  return 0;
}
long long binS(vector<long long> arr, long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binS(arr, l, mid - 1, x);
    return binS(arr, mid + 1, r, x);
  }
  return -1;
}
long long nck(long long n, long long k) {
  long long ans = 1;
  for (long long i = n; i > max(k, n - k); i--) {
    ans *= i;
  }
  for (long long i = 2; i <= min(k, n - k); i++) ans /= i;
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long fact(long long n) {
  long long ans = 1;
  for (long long i = 2; i <= n; i++) ans *= i;
  return ans;
}
long long mult(long long a, long long b) {
  long long ans = 1;
  while (b--) ans *= a;
  return ans;
}

```