#include <bits/stdc++.h>
using namespace std;
bool prime[31700 + 5];
vector<int> primes;
void sieve() {
  for (int i = 0; i < 31700 + 5; i++) prime[i] = true;
  prime[0] = prime[1] = false;
  for (int i = 2; i < 31700 + 5; i++) {
    if (prime[i] == false) continue;
    for (int j = i * i; j < 31700 + 5; j = j + i) {
      prime[j] = false;
    }
  }
  for (int i = 2; i < 31700 + 5; i++) {
    if (prime[i] == true) primes.push_back(i);
  }
}
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < (31700); i++) {
    if ((n - primes[i] - 1) % primes[i] != 0) {
      cout << primes[i] << " " << n - primes[i] - 1 << " 1\n";
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sieve();
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
bool prime[31700 + 5];
vector<int> primes;
void sieve() {
  for (int i = 0; i < 31700 + 5; i++) prime[i] = true;
  prime[0] = prime[1] = false;
  for (int i = 2; i < 31700 + 5; i++) {
    if (prime[i] == false) continue;
    for (int j = i * i; j < 31700 + 5; j = j + i) {
      prime[j] = false;
    }
  }
  for (int i = 2; i < 31700 + 5; i++) {
    if (prime[i] == true) primes.push_back(i);
  }
}
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < (31700); i++) {
    if ((n - primes[i] - 1) % primes[i] != 0) {
      cout << primes[i] << " " << n - primes[i] - 1 << " 1\n";
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sieve();
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```