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>
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
import bisect

def divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)

    divisors.sort()
    return divisors

n, m, k = map(int, input().split())
a = [int(item) for item in input().split()] + [0]
b = [int(item) for item in input().split()] + [0]

acnt = []
bcnt = []
c = 0
for item in a:
    if item == 1:
        c += 1
    else:
        acnt.append(c)
        c = 0
c = 0
for item in b:
    if item == 1:
        c += 1
    else:
        bcnt.append(c)
        c = 0
acnt.sort()
bcnt.sort()

ans = 0
div = divisors(k)
for a_val in div:
    b_val = k // a_val
    a_idx = bisect.bisect_left(acnt, a_val)
    b_idx = bisect.bisect_left(bcnt, b_val)
    a_total = sum(acnt[a_idx:]) - (len(acnt) - a_idx) * (a_val - 1)
    b_total = sum(bcnt[b_idx:]) - (len(bcnt) - b_idx) * (b_val - 1)
    ans += a_total * b_total

print(ans)