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 compress
  from math import ceil, gcd
   
  def half_sieve(n):
      if n <= 2:
          return []
      sieve = bytearray([True]) * (n // 2)
      for i in range(3, int(n ** 0.5) + 1, 2):
          if sieve[i // 2]:
              sieve[i * i // 2::i] = bytearray((n - i * i - 1) // (2 * i) + 1)
      primes = list(compress(range(1, n, 2), sieve))
      primes[0] = 2
      return primes
  
  x = half_sieve(50)
  t = int(input())
  for i in range(0,t):
      n = int(input())
      #Div by 2 because only need half of the inserted number
      
      for j in range(n):
          a = x[j]
          b = n-a-1
          if a>=b:
              break        
          
          c = gcd(a,b)
          if a!=c and b!=c and (a+b+c)==n:
              print(a,b,c)
              break