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 bisect
import itertools
import math
from functools import reduce
from itertools import accumulate
from operator import mul


def conv(x):
    n = len(x)
    i = 0
    res = []
    while i < n:
        if x[i]:
            j = 0
            while i < n and x[i]:
                i += 1
                j += 1
            res.append(j)
        else:
            i += 1
    return sorted(res)


def decomp(x):
    res = []

    while x % 2 == 0:
        res.append(2)
        x //= 2

    sqr = int(math.sqrt(x)) + 2
    i = 3
    while x > 1 and i < sqr:
        while x % i == 0:
            res.append(i)
            x //= i
        i += 2
    if x > 1:
        res.append(x)
    return res


def divs(x):
    divs = set()
    pr = decomp(x)
    for i in range(0, len(pr) + 1):
        divs.update(reduce(mul, comb, 1) for comb in itertools.combinations(pr, i))
    return divs


def go():
    n, m, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    a = conv(a)
    b = conv(b)
    n = len(a)
    m = len(b)
    ac = [0] + list(accumulate(a))
    bc = [0] + list(accumulate(b))
    res = 0

    kdivs = sorted(divs(k))

    ia = 0
    ib = m - 1
    for ka in kdivs:
        kb = k // ka
        while ia < n and a[ia] < ka:
            ia += 1
        if ia == n:
            break

        while ib >= 0 and b[ib] >= kb:
            ib -= 1

        res += (ac[-1]-ac[ia]-(n-ia)*(ka-1))*(bc[-1]-bc[ib+1]-(m-ib-1)*(kb-1))

    # i=0
    # for aa in a:
    #     while i<m and b[i]*aa>=k:
    #        i+=1
    #     res+=bc[i]-i*1

    return res


print(go())
