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 sys
  import os.path
   
  if(os.path.exists('input.txt')) :
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
      sys.stderr = open("error.txt", "w")
      
  depth = 200001
  mod = 1000000007 
  lim = mod * mod
  sys.setrecursionlimit(depth) 
   
  linp = lambda: list(minp())
  minp = lambda: map(int, input().split())
   
  from math import inf, ceil, sqrt, log2, gcd
  from collections import defaultdict, deque
  
  dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]
  
  def expo(a, b) :
      if not b : return 1
      v = expo(a, b//2)
      v = (v * v) % mod
      if b%2 : v = (v * a) % mod
      return v
  
  def modinv(a, b) :
      d = expo(b, mod-2)
      v = (a * d) % mod
      return v
  
  def checkPrime(n) :
      if n < 4 : return True
      if n%2 == 0 : return False
      for i in range(3, int(sqrt(n))+1, 2) :
          if n%i == 0 :
              return False
      return True    
  
  def highestPowerof2(x):
      x |= x >> 1
      x |= x >> 2
      x |= x >> 4
      x |= x >> 8
      x |= x >> 16
  
      return x ^ (x >> 1)
  
  p = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
  
  def findF(n) :
      for e in p :
          if n%e :
              return e
  
      return -1
  
  for _ in range(int(input())) :
      n = int(input())
      v = findF(n-1)
      print(v, n-v-1, 1)