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>
n, m, k = map(int, input().split())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))

al, bl = [], []

tmp = 0
for ai in a:
    if ai == 1:
        tmp += 1
    else:
        if tmp > 0:
            al.append(tmp)
            tmp = 0
if tmp > 0:
    al.append(tmp)

tmp = 0
for bi in b:
    if bi == 1:
        tmp += 1
    else:
        if tmp > 0:
            bl.append(tmp)
            tmp = 0
if tmp > 0:
    bl.append(tmp)


ans = 0

for i in range(1, k+1):
    if i*i > k:
        break

    if k % i != 0:
        continue

    j = k // i

    p, q, r, s = 0, 0, 0, 0
    for pi in al:
        p += max(0, pi - i + 1)
        r += max(0, pi - j + 1)

    for qi in bl:
        q += max(0, qi - j + 1)
        s += max(0, qi - i + 1)

    if i == j:
        ans += p * q
    else:
        ans += p*q + r*s

print(ans)
