Vasya has got three integers n, m and k. He'd like to find three integer points (x_1, y_1), (x_2, y_2), (x_3, y_3), such that 0 ≤ x_1, x_2, x_3 ≤ n, 0 ≤ y_1, y_2, y_3 ≤ m and the area of the triangle formed by these points is equal to nm/k.

Help Vasya! Find such points (if it's possible). If there are multiple solutions, print any of them.

Input

The single line contains three integers n, m, k (1≤ n, m ≤ 10^9, 2 ≤ k ≤ 10^9).

Output

If there are no such points, print "NO".

Otherwise print "YES" in the first line. The next three lines should contain integers x_i, y_i — coordinates of the points, one point per line. If there are multiple solutions, print any of them.

You can print each letter in any case (upper or lower).

Examples

Input

4 3 3


Output

YES
1 0
2 3
4 1


Input

4 4 7


Output

NO

Note

In the first example area of the triangle should be equal to nm/k = 4. The triangle mentioned in the output is pictured below: 

<image>

In the second example there is no triangle with area nm/k = 16/7.
import sys
from collections import defaultdict as di
range = xrange
input = sys.stdin.readline

n,m,k = [int(x) for x in input().split()]

flip = False
if m<n:
    flip=True
    n,m=m,n


big = 10**5
P = [True]*big
P[0]=P[1]=False
for i in range(big):
    if P[i]:
        j = i*i
        while j<big:
            P[j]=False
            j+=i
P = [p for p in range(big) if P[p]]

N = n
M = m
K = k

factors = di(int)
factors[2]+=1

for p in P:
    while N%p==0:
        factors[p]+=1
        N//=p
if N>1:
    factors[N]+=1

for p in P:
    while M%p==0:
        factors[p]+=1
        M//=p
if M>1:
    factors[M]+=1

for p in P:
    while K%p==0:
        factors[p]-=1
        K//=p
if K>1:
    factors[K]-=1

if any(factors[p]<0 for p in factors):
    print 'NO'
    sys.exit()

facs = {1}
for p in factors:
    for _ in range(factors[p]):
        facs |= {p*f for f in facs if p*f<=n}
facs = sorted(facs)

i = -1
while i+1<len(facs) and facs[i+1]<=n:i+=1
if i==-1:
    print 'NO'
    sys.exit()

if ((2*n*m)//k)//facs[i] <= m:
    print 'YES'
    x = facs[i]
    y = ((2*n*m)//k)//x
    print 0,0
    if flip:
        x,y=y,x
    print x,0
    print 0,y
else:
    print 'NO'
