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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.io.*;
  import java.util.*;
  
  public class B_GCD_Problem
  {
      public static void main(String[] args) 
      {
  		FastScanner fs=new FastScanner();
          int testCase=1;
          testCase=fs.nextInt();
          // int k=1;
          for(int TT=0;TT<testCase;TT++)
          {
              int n=fs.nextInt();
              int c=1;
              int a;
              int b;
              if((n-1)%2==0)
              {
                  int j=3;
                  for(;gcd((n-1-j),j)!=1;j++);
                  a=j;
                  b=n-1-j;
              }
              else
              {
                  a=2;
                  b=n-1-a;
              }
              out.println(a+" "+b+" "+c);
              // out.println("Case #" + k + ": ");
              // k++;
          }
      }
      public static int gcd(int a, int b)
      {
          if (a == 0)
              return b;
            
          return gcd(b%a, a);
      }
      static void sort(int[] a) 
      {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
      static class FastScanner 
      {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          private char NC = (char) 0;
          private char c = NC;
          private double cnt = 1;
          String next() 
          {
              while (!st.hasMoreTokens())
                  try 
                  {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) 
                  {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
          
          int nextInt() 
          {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) 
          {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() 
          {
              return Long.parseLong(next());
          }
          public double nextDouble() {
              double cur = nextLong();
              return c != '.' ? cur : cur + nextLong() / cnt;
          }
       
          public double[] nextDoubles(int N) {
              double[] res = new double[N];
              for (int i = 0; i < N; i++) {
                  res[i] = nextDouble();
              }
              return res;
          }
      }
      public static void print(int[] arr)
      {
          //for debugging only
          for(int x: arr)
              out.print(x+" ");
          out.println();
      }
      
      
  }