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.*;
import java.lang.*;
import java.io.*;


public class bgcd {
    public static void main(String[] args) throws java.lang.Exception {

        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter ino = new PrintWriter(System.out, true);
        int t = Integer.parseInt(in.readLine().trim());

        for (int i = 1; i <= t; i++) {
            int n = Integer.parseInt(in.readLine().trim());
            int a = 1, b = 2, c = n - 3;
            while (gcd(b, c) != a) {
                b++;
                c--;
            }
            ino.print(b + " " + c + " " + a + "\n");
        }
        ino.close();
    }

    static int gcd(int b, int c) {
        // int temp=Math.max(b,c);
        while (b > 1) {
            if (c % b == 0)
                return b;
            b = c % b;
        }
        return 1;
    }
}
