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 math


def handle(a: list, s: list):
    t = []
    cnt = 0
    for i in a:
        if i == 1:
            cnt += 1
        elif cnt != 0:
            t.append(cnt)
            cnt = 0
    if cnt != 0:
        t.append(cnt)
    a.clear()
    a += t
    a.sort()
    s.clear()
    s.append(0)
    for i in a:
        s.append(s[-1]+i)


def lower_bound(a: list, x: int) -> int:
    if len(a) == 0 or a[-1] < x:
        return len(a)
    l = 0
    r = len(a) - 1
    while l < r:
        m = (l+r) >> 1
        if a[m] < x:
            l = m + 1
        else:
            r = m
    return r


def cal(a: list, x: int, s: list) -> int:
    p = lower_bound(a, x)
    return s[-1] - s[p] - (x-1)*(len(a)-p)


n, m, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
s1 = []
s2 = []

handle(a, s1)
handle(b, s2)

if __name__ == '__main__':
    ans = 0
    sq = math.sqrt(k)
    i = 1
    while i < sq:
        if k % i == 0:
            ans += cal(a, i, s1)*cal(b, k//i, s2)
            ans += cal(a, k//i, s1)*cal(b, i, s2)
        i += 1
    sq = int(sq)
    if sq*sq == k:
        ans += cal(a, sq, s1)*cal(b, sq, s2)
    print(ans)