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
  input = sys.stdin.readline
  def iinput(): return int(input())
  def sinput(): return input().rstrip()
  def i0input(): return int(input()) - 1
  def linput(): return list(input().split())
  def liinput(): return list(map(int, input().split()))
  def miinput(): return map(int, input().split())
  def li0input(): return list(map(lambda x: int(x) - 1, input().split()))
  def mi0input(): return map(lambda x: int(x) - 1, input().split())
  INF = 10**20
  MOD = 1000000007
  
  from math import gcd
  
  t = iinput()
  for _ in [0] * t:
      n = iinput()
      tmp = 2
      while True:
          res = gcd(n - tmp - 1, tmp)
          if res == 1:
              print(n - tmp - 1, tmp, 1)
              break
          tmp += 1
  