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 bisect import bisect_left


def get_len(a):
    i = 0
    ret = []
    while i < len(a):
        if a[i] == 0:
            i += 1
        else:
            j = i
            while j < len(a) and a[j] == 1:
                j += 1
            ret.append(j - i)
            i = j
    return sorted(ret)


def get_sum(a):
    ret = [0]
    s = 0
    for v in a:
        s += v
        ret.append(s)
    return ret


def count_len(i, a, sa):
    p = bisect_left(a, i)
    n = len(a)
    if p == n:
        return 0
    return sa[n] - sa[p] - (i-1)*(n-p)


def solve(n, m, k, a, b):
    a, b = get_len(a), get_len(b)
    sa, sb = get_sum(a), get_sum(b)
    ret = 0
    size = []
    for i in range(1, int(k**0.5) + 1):
        if k % i == 0:
            size += [i] if k // i == i else [i, k // i]
    for i in size:
        j = k // i
        ret += count_len(i, a, sa) * count_len(j, b, sb)
    return ret


def main():
    inp = lambda: [int(x) for x in input().split()]
    n, m, k = inp()
    a, b = inp(), inp()
    print(solve(n, m, k, a, b))


if __name__ == '__main__':
    main()
