An agent called Cypher is decrypting a message, that contains a 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 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 $t$ $(1 \le t \le 100)$ — 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 $n$ $(4 \le n \le 10^9)$ — the number from the message.

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


-----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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'from sys import stdin\ninput = stdin.readline\nq = int(input())\nfor _ in range(q):\n    n = int(input())\n    pier = []\n    i = 2\n    nn = n\n    while True:\n        if nn%i == 0:\n            pier.append(i)\n            nn//=i\n        else:\n            i+= 1\n        if i**2 > n:\n            break\n    if nn != 1:\n        pier.append(nn)\n    pier_unique = list(set(pier))\n    dzielniki = [1]\n    for p in pier_unique:\n        pot = p\n        addition = []\n        while n%pot == 0:\n            addition1 = [d*pot for d in dzielniki]\n            addition += addition1\n            pot *= p\n        dzielniki += addition\n    dzielniki = dzielniki[1:]\n    k = len(pier_unique)\n    if k == 1:\n        print(*dzielniki)\n        print(0)\n    elif k >= 3:\n        dzielniki = set(dzielniki)\n        odp = []\n        for i in range(k):\n            dzielniki.remove(pier_unique[i-1]*pier_unique[i])\n        for i in range(k):\n            odp.append(pier_unique[i-1]*pier_unique[i])\n            to_rem = []\n            for dz in dzielniki:\n                if dz%pier_unique[i] == 0:\n                    to_rem.append(dz)\n                    odp.append(dz)\n            for to in to_rem:\n                dzielniki.remove(to)\n        print(*odp)\n        print(0)\n    else:\n        p = pier_unique[0]\n        q = pier_unique[1]\n        if n == p*q:\n            print(p,q,p*q)\n            print(1)\n        else:\n            test = p**2\n            if n%test != 0:\n                p,q= q,p\n            #p^2 dzieli n\n            dzielniki = set(dzielniki)\n            dzielniki.remove(p*q)\n            dzielniki.remove(p*p*q)\n            odp = [p*q]\n            to_rem = []\n            for dzu in dzielniki:\n                if dzu%p == 0:\n                    to_rem.append(dzu)\n                    odp.append(dzu)\n            for tu in to_rem:\n                dzielniki.remove(tu)\n            odp.append(p*p*q)\n            for dzu in dzielniki:\n                odp.append(dzu)\n            print(*odp)\n            print(0)\n\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14405