#include <bits/stdc++.h>
using namespace std;
long long MINll = 1e9;
struct stct {};
void sieve_of_prime(int n, vector<int>* prime) {
  bool mark[n + 1];
  memset(mark, true, sizeof(mark));
  mark[0] = mark[1] = false;
  for (int i = 2; i * i <= n; i++)
    if (mark[i])
      for (int j = 2 * i; j <= n; j += i) mark[j] = false;
  for (int i = 2; i <= n; i++)
    if (mark[i]) prime->push_back(i);
}
unsigned long long P = 1000000000 + 7;
long long MAX = -1000000000 - 1;
vector<int> prime;
unsigned long long power(unsigned long long a, unsigned long long b) {
  unsigned long long res = 1;
  a = a % P;
  if (a == 0) return 0;
  while (b) {
    if (b % 2 == 1) res = (res * a) % P;
    a = (a * a) % P;
    b = b >> 1;
  }
  return res;
}
unsigned long long bsearch(unsigned long long l, unsigned long long r,
                           unsigned long long x, unsigned long long k) {
  if (l >= r) return l;
  unsigned long long m = (l + r) / 2;
  unsigned long long sum =
      m > k ? k * k - (2 * k - m) * (2 * k - m - 1) / 2 : m * (m + 1) / 2;
  if (sum < x)
    return bsearch(m + 1, r, x, k);
  else
    return bsearch(l, m, x, k);
}
void solve() {
  int n;
  cin >> n;
  int a = 1, b = 1, c = 1;
  n--;
  a = b = n / 2;
  if (n % 2 == 0) {
    if (n % 4 == 0) {
      a--;
      b++;
    } else {
      a -= 2;
      b += 2;
    }
  } else
    b++;
  cout << a << " " << b << " " << c << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve_of_prime(1e5, &prime);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate a cpp solution 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;
long long MINll = 1e9;
struct stct {};
void sieve_of_prime(int n, vector<int>* prime) {
  bool mark[n + 1];
  memset(mark, true, sizeof(mark));
  mark[0] = mark[1] = false;
  for (int i = 2; i * i <= n; i++)
    if (mark[i])
      for (int j = 2 * i; j <= n; j += i) mark[j] = false;
  for (int i = 2; i <= n; i++)
    if (mark[i]) prime->push_back(i);
}
unsigned long long P = 1000000000 + 7;
long long MAX = -1000000000 - 1;
vector<int> prime;
unsigned long long power(unsigned long long a, unsigned long long b) {
  unsigned long long res = 1;
  a = a % P;
  if (a == 0) return 0;
  while (b) {
    if (b % 2 == 1) res = (res * a) % P;
    a = (a * a) % P;
    b = b >> 1;
  }
  return res;
}
unsigned long long bsearch(unsigned long long l, unsigned long long r,
                           unsigned long long x, unsigned long long k) {
  if (l >= r) return l;
  unsigned long long m = (l + r) / 2;
  unsigned long long sum =
      m > k ? k * k - (2 * k - m) * (2 * k - m - 1) / 2 : m * (m + 1) / 2;
  if (sum < x)
    return bsearch(m + 1, r, x, k);
  else
    return bsearch(l, m, x, k);
}
void solve() {
  int n;
  cin >> n;
  int a = 1, b = 1, c = 1;
  n--;
  a = b = n / 2;
  if (n % 2 == 0) {
    if (n % 4 == 0) {
      a--;
      b++;
    } else {
      a -= 2;
      b += 2;
    }
  } else
    b++;
  cout << a << " " << b << " " << c << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve_of_prime(1e5, &prime);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```