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.
1
669278611
 10
10
52
91
23
44
99
28
54
99
51

334639303 334639307 1
 4 5 1
25 26 1
43 47 1
9 13 1
21 22 1
47 51 1
13 14 1
26 27 1
47 51 1
23 27 1

#include <bits/extc++.h>
#include <bits/stdc++.h>

using namespace std;
using namespace __gnu_pbds;

const int maxn = 2e5 + 7;
const int maxm = 2e6 + 7;
const int mod = 998244353;

long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }

int T;
int n;
int a, b, c;

bool check() {
  return a > 0 && b > 0 && c > 0 && a + b + c == n && gcd(a, b) == c &&
         a != b && a != c && b != c;
}

int main() {
#ifndef ONLINE_JUDGE
  freopen("1.in", "r", stdin);
  //  freopen("1.out", "w", stdout);
#endif
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);

    for (c = 1; c * c <= n; c++) {
      if (n % c) {
        continue;
      }
      int num = n / c - 1;
      if (num & 1) {
        a = 2 * c;
        b = (num - 2) * c;
      } else {
        a = 3 * c;
        b = (num - 3) * c;
      }
      if (check()) {
        break;
      }
    }
    printf("%d %d %d\n", a, b, c);
  }
  return 0;
}
