human
Can you provide a Python program to find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that given the following:
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.

In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.

A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.

Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.

In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.

It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.

If there are different possible orders with a correct answer, print any of them.

Example

Input

3
6
4
30

Output

2 3 6 
1
2 4 
0
2 30 6 3 15 5 10 
0

Note

In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.

In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.

In the third test case all divisors of 30 greater than 1...
Using python3 can you solve the prior task?
assistant
#Here is Python program to find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.

from sys import stdin, stdout
    
class Input:
    def __init__(self):
        self.lines = stdin.read().splitlines()
        self.idx = 0
    
    def line(self):
        try:
            return self.lines[self.idx]
        finally:
            self.idx += 1
    
    def array(self, sep = ' ', cast = int):
        return list(map(cast, self.line().split(sep = sep)))
    
    def known_tests(self):
        num_of_cases, = self.array()
    
        for case in range(num_of_cases):
            yield self
    
    def unknown_tests(self):
        while self.idx < len(self.lines):
            yield self

def problem_solver():
    oo = float("inf")
    
    def gcd(a, b):
        return a if b == 0 else gcd(b, a % b)
    
    def modpow(a, n, m):
        ans = 1
    
        while n:
            if n & 1:
                ans = (ans * a) % m
            n >>= 1
            a = (a * a) % m
    
        return ans

    prime_numbers = []
    def primes():
        for p in prime_numbers:
            yield p

        q = prime_numbers[-1] + 1 if len(prime_numbers) else 2
        while True:
            is_prime = True
            for p in prime_numbers:
                if q % p == 0:
                    is_prime = False
                    break
            
            if is_prime:
                prime_numbers.append(q)
                yield q
            
            q += 1

    def prime_factors(n):
        factors = []

        for p in primes():
            if p * p > n:
                break
                
            if n % p != 0:
                continue

            r = 0
            while n % p == 0:
                r += 1
                n //= p

            factors.append((p, r))
        
        if n > 1:
            factors.append((n, 1))

        return factors

    '''
    '''
    def solver(n):
        f = prime_factors(n)

        divs = [1]
        e = [0, 1]
        for p, r in f:
            q = 1
            for i in range(r):
                q *= p
                for d in divs[:e[-1]]:
    ...