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 bisect import bisect_left as bl
n,m,k=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a1=[0]*n
b1=[0]*m

ans=0
a1[0]=a[0]
for i in range(1,n):
    if a[i]==1:
        a1[i]=a1[i-1]+1

b1[0]=b[0]
for i in range(1,m):
    if b[i]==1:
        b1[i]=b1[i-1]+1

a1.sort()
b1.sort()

ans=0
for i in range(1,int(k**0.5)+1):
    if k%i==0:
        if i==k//i:
            x1=n-bl(a1,i)
            x2=m-bl(b1,i)
            ans+=x1*x2
        else:
            x1=n-bl(a1,i)
            x2=m-bl(b1,k//i)
            ans+=x1*x2

            x1=n-bl(a1,k//i)
            x2=m-bl(b1,i)
            ans+=x1*x2
print(ans)