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

import sys
import os.path
 
if(os.path.exists('input.txt')) :
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("output.txt", "w")
    sys.stderr = open("error.txt", "w")
    
depth = 200001
mod = 1000000007 
lim = mod * mod
sys.setrecursionlimit(depth) 
 
linp = lambda: list(minp())
minp = lambda: map(int, input().split())
 
from math import inf, ceil, sqrt, log2, gcd
from collections import defaultdict, deque

dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]

def expo(a, b) :
    if not b : return 1
    v = expo(a, b//2)
    v = (v * v) % mod
    if b%2 : v = (v * a) % mod
    return v

def modinv(a, b) :
    d = expo(b, mod-2)
    v = (a * d) % mod
    return v

def checkPrime(n) :
    if n < 4 : return True
    if n%2 == 0 : return False
    for i in range(3, int(sqrt(n))+1, 2) :
        if n%i == 0 :
            return False
    return True    

def highestPowerof2(x):
    x |= x >> 1
    x |= x >> 2
    x |= x >> 4
    x |= x >> 8
    x |= x >> 16

    return x ^ (x >> 1)

p = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]

def findF(n) :
    for e in p :
        if n%e :
            return e

    return -1

for _ in range(int(input())) :
    n = int(input())
    v = findF(n-1)
    print(v, n-v-1, 1)