#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int limit = 5e5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
const int mod = 1e9 + 7;
const int M = 500 + 5;
const int S = (1 << 13) + 5;
const int N = 2e5 + 5;
int n, m, _, k;
long long a[N];
long long b[N], d[N];
signed main() {
  int T;
  while (cin >> T)
    while (T--) {
      scanf("%d", &n);
      if (n & 1) {
        int mid = n >> 1;
        if (mid & 1)
          printf("%d %d %d\n", n / 2 - 2, n / 2 + 2, 1);
        else
          printf("%d %d %d\n", n / 2 - 1, n / 2 + 1, 1);
      } else {
        printf("%d %d %d\n", n / 2, n / 2 - 1, 1);
      }
    }
  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;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int limit = 5e5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
const int mod = 1e9 + 7;
const int M = 500 + 5;
const int S = (1 << 13) + 5;
const int N = 2e5 + 5;
int n, m, _, k;
long long a[N];
long long b[N], d[N];
signed main() {
  int T;
  while (cin >> T)
    while (T--) {
      scanf("%d", &n);
      if (n & 1) {
        int mid = n >> 1;
        if (mid & 1)
          printf("%d %d %d\n", n / 2 - 2, n / 2 + 2, 1);
        else
          printf("%d %d %d\n", n / 2 - 1, n / 2 + 1, 1);
      } else {
        printf("%d %d %d\n", n / 2, n / 2 - 1, 1);
      }
    }
  return 0;
}

```