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.
def factorization(n):
    ans = []
    while (n % 2 == 0):
        n //= 2
        yield 2
    d = 3
    while d * d <= n:
        while n % d == 0:
            n //= d
            yield d
        d += 2
    if n > 1:
        yield n


def solve1(n, m, k):
    kmod2 = False
    if k % 2 == 0:
        k //= 2
        kmod2 = True
    x, y = n, m
    ok = True
    for d in factorization(k):
        if x % d == 0:
            x //= d
        elif y % d == 0:
            y //= d
        else:
            ok = False
            break
    if not kmod2:
        if x * 2 <= n:
            x *= 2
        elif y * 2 <= m:
            y *= 2
        else:
            ok = False

    if ok:
        print('YES')
        print(0, 0)
        print(x, 0)
        print(0, y)
    return ok


def solve2(n, m, k):
    kmod2 = False
    if k % 2 == 0:
        k //= 2
        kmod2 = True
    x, y = n, m
    ok = True
    for d in factorization(k):
        if y % d == 0:
            y //= d
        elif x % d == 0:
            x //= d
        else:
            ok = False
            break
    if not kmod2:
        if x * 2 <= n:
            x *= 2
        elif y * 2 <= m:
            y *= 2
        else:
            ok = False

    if ok:
        print('YES')
        print(0, 0)
        print(x, 0)
        print(0, y)
    return ok


n, m, k = map(int, input().split())

if (2 * n * m) % k != 0:
    print('NO')
else:
    if not solve1(n, m, k) and \
       not solve2(n, m, k):
        print('NO')
