#include <bits/stdc++.h>
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC optimize("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-funroll-all-loops,-fpeel-loops,-funswitch-loops")
int dy[] = {1, -1, 0, 0};
int dx[] = {0, 0, 1, -1};
using namespace std;
using namespace chrono;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (int i = 2;; i++) {
      if (gcd(i, (n - i - 1)) == 1) {
        cout << i << ' ' << n - i - 1 << ' ' << 1 << '\n';
        break;
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC optimize("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-funroll-all-loops,-fpeel-loops,-funswitch-loops")
int dy[] = {1, -1, 0, 0};
int dx[] = {0, 0, 1, -1};
using namespace std;
using namespace chrono;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (int i = 2;; i++) {
      if (gcd(i, (n - i - 1)) == 1) {
        cout << i << ' ' << n - i - 1 << ' ' << 1 << '\n';
        break;
      }
    }
  }
  return 0;
}

```