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

import java.util.*;
public class perform
{
    public static void main(String args[])
{
Scanner in=new Scanner(System.in);
int t=0,n=0,a=0,k=0;
t=in.nextInt();
for(int p=1;p<=t;p++)
{
n=in.nextInt();
abc:for(int i=1;;i++)
{
 a=n-i;
 for(int j=i+1;j<=(a-1)/2;j++)
 {
    k=a-j;
    if(i==j||j==k||i==k);
    else
    {
    if(gcd(i,j)==k)
    {
      System.out.println(i+" "+j+" "+k);
      break abc;
    }
    else if(gcd(k,j)==i)
    {System.out.println(k+" "+j+" "+i);break abc;}
    else if(gcd(i,k)==j)
    {System.out.println(i+" "+k+" "+j);break abc;}
    }
}
}}}
public static int gcd(int a,int b)
 {  
    int v=0,g=1;
    if(a<b)
    v=a;
else
v=b;
for(int i=2;i<=v;i++)
{
if(a%i==0 && b%i==0)
g=i;}
return g;}}