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/stdc++.h>
using namespace std;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
int mod_add(int a, int b, int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
int mod_mul(int a, int b, int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
int mod_sub(int a, int b, int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
const long long mxm = 2 * 1e5;
long long M = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
long long expo(long long a, long long b, long long m) {
  if (b == 0) {
    return 1;
  }
  if (b % 2 == 0) {
    long long temp = expo(a, b / 2, m);
    return (temp % m * temp % m) % m;
  } else {
    long long temp = expo(a, b / 2, m);
    return (a % m * temp % m * temp % m) % m;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long c = 1;
    long long a, b;
    if ((n - 1) % 2 == 0) {
      a = (n - 1) / 2 + 1;
      b = (n - 1) / 2 - 1;
      if (a % 2 == 0) {
        c = 3;
        a = (n - 3) / 2 + 1;
        b = (n - 3) / 2 - 1;
      }
    } else {
      a = (n - 1) / 2;
      b = (n - 1) - (n - 1) / 2;
    }
    cout << a << ' ' << b << ' ' << c << "\n";
  }
  return 0;
}
