description: 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.
solution: 
  
  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;
      }
  }
  