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

# Plan
# Get all factorisations of k of the form k = x * y
# Find number of runs in a of size x and the number of runs in b of size y
# If x != y, repeat swapping a and b
# Take the product of these run numbers


def get_factor_pairs(x):
    out = []
    for i in range(1, 1 + int(math.sqrt(x))):
        if x % i == 0:
            out.append([i, x//i])
    return out


def get_seq(arr, arr_length):
    out = []
    for i in range(arr_length):
        if arr[i] == 0:
            out.append(0)
        elif arr[i] == 1 and i == 0:
            out.append(1)
        else:
            out.append(out[i-1]+1)
    return out

def get_count(arr, val):
    count = 0
    for i in arr:
        if i >= val:
            count += 1
    return count


n, m, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
pairs = get_factor_pairs(k)
a_seq = get_seq(a, n)
b_seq = get_seq(b, m)

total = 0
for x, y in pairs:
    a_count = get_count(a_seq, x)
    b_count = get_count(b_seq, y)
    total += a_count * b_count

    if x != y:
        a_count = get_count(a_seq, y)
        b_count = get_count(b_seq, x)
        total += a_count * b_count

print(total)






