#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  return a.second < b.second;
}
string bin(long long int n) {
  string s = "";
  while (n) {
    if (n % 2)
      s += '1';
    else
      s += '0';
    n /= 2;
  }
  reverse(s.begin(), s.end());
  return s;
}
void Solve() {
  long long int n;
  cin >> n;
  long long int a, b, c;
  if ((n - 1) % 2) {
    c = 1;
    a = 2;
    b = n - 3;
    cout << a << " " << b << " " << c << '\n';
    return;
  }
  c = 1;
  long long int x = (n - 1) / 2;
  if (x % 2) {
    a = x - 2;
    b = x + 2;
  } else {
    a = x - 1;
    b = x + 1;
  }
  cout << a << " " << b << " " << c << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long int T;
  cin >> T;
  while (T--) Solve();
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
bool comp(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  return a.second < b.second;
}
string bin(long long int n) {
  string s = "";
  while (n) {
    if (n % 2)
      s += '1';
    else
      s += '0';
    n /= 2;
  }
  reverse(s.begin(), s.end());
  return s;
}
void Solve() {
  long long int n;
  cin >> n;
  long long int a, b, c;
  if ((n - 1) % 2) {
    c = 1;
    a = 2;
    b = n - 3;
    cout << a << " " << b << " " << c << '\n';
    return;
  }
  c = 1;
  long long int x = (n - 1) / 2;
  if (x % 2) {
    a = x - 2;
    b = x + 2;
  } else {
    a = x - 1;
    b = x + 1;
  }
  cout << a << " " << b << " " << c << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long int T;
  cin >> T;
  while (T--) Solve();
  return 0;
}

```