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 
n,m,k=map(int,input().split())
if (2*n*m)%k==0:
    s=set()
    f=0
    for i in range(1,int(n**0.5)+1):
        if n%i==0:
            s.add(i)
            s.add(n//i)
    s.add(n)
    s1=set()
    for i in range(2,int(m**0.5)+1):
        if m%i==0:
            s1.add(i)
            s1.add(m//i)
        #print(s1)
    s1.add(m)
    if m==1:
        if n%k==0:
            print("YES")
            print(0,0)
            print(2*n//k,0)
            print(0,1)
            sys.exit()
    elif n==1:
        if m%k==0:
            print("YES")
            print(0,0)
            print(1,0)
            print(0,2*m//k)
            sys.exit()
    #print(s,s1)
    for i in s:
        if k%i==0 and k//i in s1:
            print("YES")
            print(0,0)
            print(n//i,0)
            print(0,(2*m)//(k//i))
            f=1
            break
        elif k%(2*i)==0 and (k//(2*i) in s1 or k==2*i):
            print("YES")
            print(0,0)
            print(n//i,0)
            print(0,(2*m)//(k//i))
            f=1
            break
    if f==0:
        s.remove(1)
        s1.add(1)
        #print(s,s1)
        for i in s1:
           if k%i==0 and k//i in s:
            print("YES")
            print(0,0)
            print(0,m//i)
            print((2*n)//(k//i),0)
            f=1
            break
           elif k%(2*i)==0 and (k//(2*i) in s or k//(2*i)==1):
            print("YES")
            print(0,0)
            print(0,m//i)
            print((2*n)//(k//i),0)
            f=1
            break
        if f==0:
            print("NO")
        
else:
    print("NO")
                
        