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.
1
669278611
 10
10
52
91
23
44
99
28
54
99
51

334639303 334639307 1
 4 5 1
25 26 1
43 47 1
9 13 1
21 22 1
47 51 1
13 14 1
26 27 1
47 51 1
23 27 1

#from itertools import *
from math import *
#from bisect import *
#from collections import *
#from random import *
#from decimal import *
#from heapq import *
#from itertools import *            # Things Change ....remember :)
import sys
input=sys.stdin.readline
def inp():
    return int(input())
def st():
    return input().rstrip('\n')
def lis():
    return list(map(int,input().split()))
def ma():
    return map(int,input().split())
t=inp()
while(t):
    t-=1
    n=inp()
    x=n-1
    y=x//2
    z=x-y
    if(n%2):
        if(gcd(3,n-4)==1):
            y=3
            z=n-4
        else:
            y=5
            z=n-6
    print(y,z,1)
