Problem: 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 can be placed in some order so that there are no two adjacent numbers that are coprime.

Solution (in PYTHON3):

Solution:
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------

import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading




BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
    
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
     
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
  
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#-------------------game starts now-----------------------------------------------------
class Factorial:
    def __init__(self, MOD):
        self.MOD = MOD
        self.factorials = [1, 1]
        self.invModulos = [0, 1]
        self.invFactorial_ = [1, 1]
 
    def calc(self, n):
        if n <= -1:
            print("Invalid argument to calculate n!")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        if n < len(self.factorials):
            return self.factorials[n]
        nextArr = [0] * (n + 1 - len(self.factorials))
        initialI = len(self.factorials)
        prev = self.factorials[-1]
        m = self.MOD
        for i in range(initialI, n + 1):
            prev = nextArr[i - initialI] = prev * i % m
        self.factorials += nextArr
        return self.factorials[n]
 
    def inv(self, n):
        if n <= -1:
            print("Invalid argument to calculate n^(-1)")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        p = self.MOD
        pi = n % p
        if pi < len(self.invModulos):
            return self.invModulos[pi]
        nextArr = [0] * (n + 1 - len(self.invModulos))
        initialI = len(self.invModulos)
        for i in range(initialI, min(p, n + 1)):
            next = -self.invModulos[p % i] * (p // i) % p
            self.invModulos.append(next)
        return self.invModulos[pi]
 
    def invFactorial(self, n):
        if n <= -1:
            print("Invalid argument to calculate (n^(-1))!")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        if n < len(self.invFactorial_):
            return self.invFactorial_[n]
        self.inv(n)  # To make sure already calculated n^-1
        nextArr = [0] * (n + 1 - len(self.invFactorial_))
        initialI = len(self.invFactorial_)
        prev = self.invFactorial_[-1]
        p = self.MOD
        for i in range(initialI, n + 1):
            prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
        self.invFactorial_ += nextArr
        return self.invFactorial_[n]
 
 
class Combination:
    def __init__(self, MOD):
        self.MOD = MOD
        self.factorial = Factorial(MOD)
 
    def ncr(self, n, k):
        if k < 0 or n < k:
            return 0
        k = min(k, n - k)
        f = self.factorial
        return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
def get_divisors(n):
    ret = set()
    i = 1
    while i <= math.sqrt(n):
        if (n % i == 0):
            if (n//i == i):
                ret.add(i)
            else:
                ret.add(i)
                ret.add(n//i)
        i += 1
    return ret
 
for _ in range(int(input())):
    n = int(input())
    o = n
    # prime factorization
    primes = {}
    while n % 2 == 0:
        if 2 in primes:
            primes[2] += 1
        else:
            primes[2] = 1
        n = n // 2
    for i in range(3,int(math.sqrt(n))+1,2):
        while n % i == 0:
            if i in primes:
                primes[i] += 1
            else:
                primes[i] = 1
            n = n // i
    if n > 2:
        primes[n] = 1
    if len(primes) == 1:
        p = list(primes.keys())[0]
        for i in range(primes[p]):
            print(p**(i+1),end=" ")
        print()
        print(0)
    elif len(primes) == 2:
        l = list(primes.keys())
        p1 = l[0]
        p2 = l[1]
        e1 = primes[p1]
        e2 = primes[p2]
        if e1 == 1 and e2 == 1:
            print(p1,p2,p1*p2)
            print(1)
        elif e1 >= e2:
            for i in range(e2):
                print(p2**(i+1),end=" ")
            print(p2*p1,end=" ")
            for i in range(1,e1+1):
                for j in range(e2+1):
                    if j == 1 and (i == 1 or i == 2):
                        continue
                    print(p1**i*p2**j,end=" ")
            print(p2*p1**2)
            print(0)
        else:
            for i in range(e1):
                print(p1**(i+1),end=" ")
            print(p2*p1,end=" ")
            for i in range(1,e2+1):
                for j in range(e1+1):
                    if j == 1 and (i == 1 or i == 2):
                        continue
                    print(p2**i*p1**j,end=" ")
            print(p1*p2**2)
            print(0)
    else:
        divisors = get_divisors(o)
        divisors.remove(1)
        ans = []
        keys = list(primes.keys())
        for k in range(len(keys)):
            p = keys[k]
            nextp = keys[(k+1)%(len(keys))]
            curr = o // p
            divs = get_divisors(curr)
            l = {p*x for x in divs}
            l.remove(p*nextp)
            if k == 0:
                l.remove(p*keys[len(keys)-1])
                divisors.remove(p*keys[len(keys)-1])
            for el in l:
                if el in divisors:
                    ans.append(el)
                    divisors.remove(el)
            ans.append(p*nextp)
            if p*nextp in divisors:
                divisors.remove(p*nextp)
        print(*ans)
        print(0)