#include <bits/stdc++.h>
using namespace std;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const long long mod = (long long)1e9 + 7;
vector<long long> primes;
void simpleSieve(long long lmt, vector<long long>& prime) {
  bool mark[lmt + 1];
  memset(mark, false, sizeof(mark));
  for (long long i = 2; i <= lmt; ++i) {
    if (!mark[i]) {
      prime.push_back(i);
      for (long long j = i; j <= lmt; j += i) {
        mark[j] = true;
      }
    }
  }
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  long long n;
  cin >> n;
  long long c = 1;
  for (auto x : primes) {
    long long a = n - x - 1;
    long long b = x;
    if (gcd(a, b) == 1) {
      cout << a << " " << b << " " << c << "\n";
      return;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  long long tt;
  tt = 1;
  simpleSieve(1000000, primes);
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

### 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;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const long long mod = (long long)1e9 + 7;
vector<long long> primes;
void simpleSieve(long long lmt, vector<long long>& prime) {
  bool mark[lmt + 1];
  memset(mark, false, sizeof(mark));
  for (long long i = 2; i <= lmt; ++i) {
    if (!mark[i]) {
      prime.push_back(i);
      for (long long j = i; j <= lmt; j += i) {
        mark[j] = true;
      }
    }
  }
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  long long n;
  cin >> n;
  long long c = 1;
  for (auto x : primes) {
    long long a = n - x - 1;
    long long b = x;
    if (gcd(a, b) == 1) {
      cout << a << " " << b << " " << c << "\n";
      return;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  long long tt;
  tt = 1;
  simpleSieve(1000000, primes);
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

```