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

# sys.stdin = open('test')


def get_dividers(k):
    dividers = set()
    for i in range(1, int(k ** 0.5) + 2):
        if k % i == 0:
            dividers.add(i)
            dividers.add(k // i)

    return sorted(list(dividers))


n, m, k = map(int, input().split())
dividers = get_dividers(k)
A = list(map(int, input().split()))
B = list(map(int, input().split()))



def count(A, d):
    in_a_row = 0
    res = 0
    for a in A:
        if a == 1:
            in_a_row += 1
            if in_a_row >= d:
                res += 1
        else:
            in_a_row = 0
    return res


a_d = {}
b_d = {}
for d in dividers:
    a_d[d] = count(A, d)
    b_d[d] = count(B, d)

res = 0
for d in dividers:
    res += a_d[d] * b_d[k//d]
    # res += b_d[d] * a_d[k//d]
    # print(d, a_d[d] * b_d[k//d], b_d[d] * a_d[k//d], res)
print(res)