#include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const long long N = 0;
void solve() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    n--;
    long long x = n / 2;
    long long y = n - x;
    cout << x << " " << y << " " << 1 << "\n";
    return;
  }
  long long f = n;
  long long p = n / f;
  f--;
  long long xx = f / 2;
  if (xx & 1) {
    cout << p * (xx - 2) << " " << (f - xx + 2) * p << " " << p << "\n";
  } else
    cout << p * (xx - 1) << " " << (xx + 1) * p << " " << p << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
}

### Prompt

Please create a solution in CPP 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 binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const long long N = 0;
void solve() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    n--;
    long long x = n / 2;
    long long y = n - x;
    cout << x << " " << y << " " << 1 << "\n";
    return;
  }
  long long f = n;
  long long p = n / f;
  f--;
  long long xx = f / 2;
  if (xx & 1) {
    cout << p * (xx - 2) << " " << (f - xx + 2) * p << " " << p << "\n";
  } else
    cout << p * (xx - 1) << " " << (xx + 1) * p << " " << p << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
}

```