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 math import *

n,m,k = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))

cna = []
cn = 0 
for i in a:
    if(i == 1):
        cn += 1
    else:
        cna.append(cn)
        cn = 0 
cna.append(cn)

cnb = []
cn = 0 
for i in b:
    if(i == 1):
        cn += 1
    else:
        cnb.append(cn)
        cn = 0 
cnb.append(cn)


da = [0]*(n+1)
db = [0]*(m+1)
for i in  cna :
    for j in range(1,i+1):
        da[j] += i-j+1
for i in cnb:
    for j in range(1,i+1):
        db[j] += i-j+1

# print(da,db)

ans = 0 
h = int(sqrt(k))
# print(h+1)
for i in range(1,h+1):
    if(k%i == 0):
        f = k//i
        if(i!=f):
            if(i<=n and f<=m):
                ans += da[i]*db[f]
            if(i<=m and f<=n):
                ans += da[f]*db[i]
        else:
            if(i>n or i>m):
                continue
            else:
                ans += da[i]*db[f]
        # print(ans,i,f)

print(ans)