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: def gcd_1(x, y):
      """greatest common divisor of x and y"""
      while y:
          x, y = y, x % y
      return x
  
  
  def extended_gcd(a, b):
      """returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
      s, old_s = 0, 1
      r, old_r = b, a
      while r:
          q = old_r // r
          old_r, r = r, old_r - q * r
          old_s, s = s, old_s - q * s
      return old_r, old_s, (old_r - old_s * a) // b if b else 0
  from math import gcd
  t = int(input())
  
  while t:
      num = int(input())
      
      first = 2
      second = num - 3
      third = 1
      
      while first < second and gcd_1(second, first) != 1:
           #check until found
          first += 1
          second -= 1
      print(first, second, third)
      t -= 1