You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.

How many subrectangles of size (area) k consisting only of ones are there in c?

A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.

The size (area) of a subrectangle is the total number of cells in it.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.

The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.

Output

Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.

Examples

Input


3 3 2
1 0 1
1 1 1


Output


4


Input


3 5 4
1 1 1
1 1 1 1 1


Output


14

Note

In first example matrix c is:

<image>

There are 4 subrectangles of size 2 consisting of only ones in it:

<image>

In second example matrix c is:

<image>
from math import sqrt
from collections import Counter

def get_parts(x):
    start = None
    parts = []
    for i, s in enumerate(x):
        if s == 0 and start is not None:
            size = i - start
            if size > 0:
                parts.append(size)
                start = None
        if s == 1 and start is None:
            start = i
    if start is not None:
        parts.append(len(x) - start)
    return parts

def get_divisors(k):
    div = {1}
    for i in range(2, int(sqrt(k)) + 1):
        if k % i == 0:
            div.add(i)
            div.add(k // i)
    div.add(k)
    return sorted(div)

def count_n(n, m):
    return n - m + 1

def solve(n, m, k, a, b):
    a_parts = Counter(get_parts(a))
    b_parts = Counter(get_parts(b))
    res = 0
    div = get_divisors(k)
    for a_part, a_count in a_parts.items():
        for b_part, b_count in b_parts.items():
            if a_part * b_part >= k:
                for i in div:
                    if i <= a_part and k // i <= b_part:
                        res += count_n(a_part, i) * count_n(b_part, k // i) * a_count * b_count

    return res


if __name__ == '__main__':
    n, m, k = [int(x) for x in input().split(" ")]
    a = [int(x) for x in input().split(" ")]
    b = [int(x) for x in input().split(" ")]
    print(solve(n, m, k, a, b))