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

/*
 author:himanshu3198

*/
#include <bits/stdc++.h>
#define ll long long int
#define fr(i, n) for (int i = 0; i < (n); i++)
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define FORR(i, n) for (int i = (n); i >= 0; i--)
#define fst first
#define snd second
#define E "\n"
#define pb push_backCF
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define MOD 1000000007LL
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define ABS(x) ((x) < 0 ? -(x) : (x))
#define setbits(x) __builtin_popcount(x)
#define fastio()                      \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);
using namespace std;

void solve()
{
   
   int n=18;
   cin>>n;
     
      int a,b,c;
     for(int i=2;i*i<=n;i++){

          a=i;
          b=n-i-1;
         c=__gcd(a,b);
         if(a!=b and a!=c  and b!=c and b!=a and  c!=a and c!=b and a+b+c==n){
             break;
         }
     }

     cout<<a<<" "<<b<<" "<<c<<E;



  
  
}


int main()
{
    fastio();
    int t = 1;
    cin >> t;

    while (t--)
    {

        solve();
    }

    return 0;
}
