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>
def add_to_dict(d, k):
    if k == 0: return
    if k in d:
        d[k] += 1
    else:
        d[k] = 1

def process(A):
    k = 0
    N1 = {}
    for a in A:
        if a == 0:
            add_to_dict(N1, k)
            k = 0
        else:
            k += 1
    add_to_dict(N1, k)
    return list(N1.items())

def find_n(N1, d):
    res = 0
    for l, n in N1:
        res += max(0, l-d+1)*n
    return res

def solve(A, B, k):
    AN1 = process(A)
    BN1 = process(B)

    res = 0
    to = min(len(A), k)
    for i in range(1, to+1):
        if k%i != 0 or k//i > len(B): continue
        res += find_n(AN1, i)*find_n(BN1, k//i)
    return res

_, _, k = list(map(int, input().split()))
A = list(map(int, input().split()))
B = list(map(int, input().split()))
print(solve(A, B, k))