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>
from _collections import  OrderedDict
def div_(num, mx):
    mas = list()
    for i in range(1, min(num, mx) + 1):
        if num % i == 0:
            mas.append((i, num // i))
    return mas


n, m, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
ans = 0
lenght = 0
row = OrderedDict()
for i in a:
    if i == 1:
        lenght += 1
    else:
        if lenght:
            if lenght in row:
                row[lenght] += 1
            else:
                row[lenght] = 1
        lenght = 0
if lenght:
    if lenght in row:
        row[lenght] += 1
    else:
        row[lenght] = 1
column = OrderedDict()
lenght = 0
for i in b:
    if i == 1:
        lenght += 1
    else:
        if lenght:
            if lenght in column:
                column[lenght] += 1
            else:
                column[lenght] = 1
        lenght = 0
if lenght:
    if lenght in column:
        column[lenght] += 1
    else:
        column[lenght] = 1
mdiv = div_(k, max(m, n))
if len(row) and len(column):
    for k in mdiv:
        x, y = [int(i) for i in k]
        for i in row:
            for j in column:
                if i >= x and j >= y:
                    ans += (i - x + 1) * (j - y + 1) * row[i] * column[j]
print(ans)