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: #from itertools import *
  from math import *
  #from bisect import *
  #from collections import *
  #from random import *
  #from decimal import *
  #from heapq import *
  #from itertools import *            # Things Change ....remember :)
  import sys
  input=sys.stdin.readline
  def inp():
      return int(input())
  def st():
      return input().rstrip('\n')
  def lis():
      return list(map(int,input().split()))
  def ma():
      return map(int,input().split())
  t=inp()
  while(t):
      t-=1
      n=inp()
      x=n-1
      y=x//2
      z=x-y
      if(n%2):
          po=3
          while(1):
              if(gcd(po,n-po-1)==1):
                  y=po
                  z=n-po-1
                  break
              po+=2
      print(y,z,1)
  