You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.

Otherwise, print "YES" and any possible such representation.

Example

Input


5
64
32
97
2
12345


Output


YES
2 4 8 
NO
NO
NO
YES
3 5 823 
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f

class Prime():
    def __init__(self, n):
        self.M = m = int(math.sqrt(n)) + 10
        self.A = a = [True] * m
        a[0] = a[1] = False
        self.T = t = [2]
        for j in range(4, m, 2):
            a[j] = False
        for i in range(3, m, 2):
            if not a[i]:
                continue
            t.append(i)
            for j in range(i*i,m,i):
                a[j] = False
        self.ds_memo = {}
        self.ds_memo[1] = set([1])

    def is_prime(self, n):
        return self.A[n]

    def division(self, n):
        d = collections.defaultdict(int)
        for c in self.T:
            while n % c == 0:
                d[c] += 1
                n //= c
            if n < 2:
                break
        if n > 1:
            d[n] += 1
        return d.items()

    # memo
    def divisions(self, n):
        if n in self.ds_memo:
            return self.ds_memo[n]

        for c in self.T:
            if n % c == 0:
                rs = set([c])
                for cc in self.divisions(n // c):
                    rs.add(cc)
                    rs.add(cc * c)
                self.ds_memo[n] = rs
                return rs

        rs = set([1, n])
        self.ds_memo[n] = rs
        return rs

    def sowa(self, n):
        r = 1
        for k,v in self.division(n):
            t = 1
            for i in range(1,v+1):
                t += math.pow(k, i)
            r *= t
        return r

def main():
    t = I()

    pr = Prime(10**10)
    rr = []
    for _ in range(t):
        n = I()

        ds = sorted(pr.division(n))
        r = []
        if len(ds) == 1:
            k,v = ds[0]
            if v < 6:
                rr.append("NO")
                continue
            r = [k, k**2, k**(v-3)]
        else:
            t = n // ds[0][0]
            t //= ds[1][0]
            if t < 2 or t == ds[0][0] or t == ds[1][0]:
                rr.append("NO")
                continue
            r = [ds[0][0], ds[1][0], t]

        rr.append("YES")
        rr.append(JA(r, " "))


    return JA(rr, "\n")


print(main())


