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
import math
from collections import Counter
from functools import lru_cache


n, m, k = map(int, next(sys.stdin).split())

a1 = list(map(int, next(sys.stdin).split()))
a2 = list(map(int, next(sys.stdin).split()))


def get_ones_list(a):
    result = []
    current_length = 0
    for x in a:
        if x == 1:
            current_length += 1
        elif x == 0 and current_length > 0:
            result.append(current_length)
            current_length = 0
    if current_length > 0:
        result.append(current_length)
    return result


@lru_cache(maxsize=10)
def get_sizes(kk):
    result = set()
    for x in range(1, int(math.sqrt(kk) + 1)):
        if kk % x == 0:
            result.add((x, int(kk / x)))
            result.add((int(kk / x), x))
    return list(result)


@lru_cache(maxsize=None)
def get_n_inners(a, b, k):
    s = 0

    if k > a * b:
        return 0
    elif k == a * b:
        return 1

    for size_x, size_y in get_sizes(k):
        if size_x > a:
            continue
        if size_y > b:
            continue
        s += (a - size_x + 1) * (b - size_y + 1)
    return s


r = 0
counter_a1 = Counter(get_ones_list(a1))
counter_a2 = Counter(get_ones_list(a2))
for aa, n1 in counter_a1.items():
    for bb, n2 in counter_a2.items():
        r += get_n_inners(aa, bb, k) * n1 * n2

print(r)
