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 math
  from collections import deque,Counter
  from sys import stdin
  
  #sys.setrecursionlimit(10**7)
  
  int1=lambda x: int(x)-1
  stinput=lambda :stdin.readline()[:-1]
  ii=lambda :int(stinput())
  mi=lambda :map(int, stdin.readline().split())
  li=lambda :list(mi())
  mi1=lambda :map(int1, stdin.readline().split())
  li1=lambda :list(mi1())
  mis=lambda :map(str, stdin.readline().split())
  lis=lambda :list(mis())
  
  from collections import defaultdict
  #d=defaultdict(int)
  #d=defaultdict(lambda:1)
  
  import heapq
  #heapq.heapify(hq)
  #heapq.heappop(hq)
  #heapq.heappush(hq,1)
  
  mod=10**9+7
  Mod=998244353
  INF=10**18
  ans=0
  
  t=ii()
  for _ in range(t):
    n=ii()
    n-=1
    if n%2==1:
      print(n//2,(n+1)//2,1)
    else:
      m=n//2
      if m%2==0:
        print(m-1,m+1,1)
      else:
        print(m-2,m+2,1)