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

def isPrime(n):
    for i in range(2, int(n**0.5)+1):
        if n % i == 0:
            return False

    return True

def gcd(x, y):
  
   while(y):
       x, y = y, x % y
   return x


t = int(input())

for i in range(t):
    n = int(input())

    if not isPrime(n):
        for i in range(1, n):
            if n % i == 0 and n-i-n/i>0:
                if gcd(i, n-i-n/i) == n/i:
                    print(i, int(n-i-n/i), int(n/i))
                    break
                else:
                    hashmap={}
                    x=i
                    m = n-x
                    for i in range(1,m):
                        done = 0
                        if gcd(i,m-i) == x:
                            print(i,int(m-i),x)
                            done = 1
                            break
                    if done: break


    else:
        hashmap = {}
        m = n-1
        for i in range(1,m):
            if gcd(i,m-i) == 1:
                print(i,int(m-i),1)
                break

