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>
import sys

input = sys.stdin.readline
import collections


def make_divisors(n):
    divisors = []
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n // i)

    divisors.sort()
    return divisors


def 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()]

    divs = make_divisors(K)

    ac = collections.Counter()
    acont = 0
    for a in A:
        if a == 0:
            for d in divs:
                if acont >= d:
                    ac[d] += acont - d + 1
                else:
                    break
            acont = 0
        else:
            acont += 1
    else:
        for d in divs:
            if acont >= d:
                ac[d] += acont - d + 1
            else:
                break
    bcont = 0
    bc = collections.Counter()
    for b in B:
        if b == 0:
            for d in divs:
                if bcont >= d:
                    bc[d] += bcont - d + 1
                else:
                    break
            bcont = 0
        else:
            bcont += 1
    else:
        for d in divs:
            if bcont >= d:
                bc[d] += bcont - d + 1
            else:
                break

    ans = 0
    for d in divs:
        ad = K // d
        ans += ac[d] * bc[ad]

    print(ans)


if __name__ == '__main__':
    main()
