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
input = sys.stdin.readline

def find_r(a,n):
    rects = []
    i = 0
    while i < n:
        cnt = 0
        while i < n and a[i] == 1:
            i+=1
            cnt += 1
        rects.append(cnt)
        while i < n and a[i] == 0:
            i+=1
    
    cnt_lst = dict()
    for val in rects:
        try:
            cnt_lst[val] += 1
        except:
            cnt_lst[val] = 1
    return cnt_lst

n,m,k = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]

recta = find_r(a,n)
rectb = find_r(b,m)
ans = 0

i = 1
factors = []
while i <= k**0.5: 
    if (k % i == 0) : 
        if (k // i == i) : 
            factors.append(i) 
        else: 
            factors.extend((i , k//i))
    i += 1
factors = sorted(factors)
lsta = [[max(0, vala*(keya-f+1)) for f in factors] for keya,vala in recta.items()]
lstb = [[max(0, valb*(keyb-f+1)) for f in factors[::-1]] for keyb,valb in rectb.items()]
for la in lsta:
    for lb in lstb:
        ans += sum((la[ii]*lb[ii] for ii in xrange(len(factors))))
print(ans)