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 ceil
n,m,k=map(int,input().split())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
d,ans,last=[],0,-1
for i in range(1,ceil(k**(1/2))+1):
    if k%i==0:
        if i==k//last:
            break
        d.append(i)
        last=i
count=0
a,b=[0]*40001,[0]*40001
for i in range(n):
    if x[i]==1:
        count+=1
    if count>0 and (x[i]==0 or i==n-1):
        p=1
        for i2 in range(count,0,-1):
            a[i2]+=p
            p+=1
        count=0
for i in range(m):
    if y[i] == 1:
        count += 1
    if count > 0 and (y[i] == 0 or i == m - 1):
        p = 1
        for i2 in range(count, 0, -1):
            b[i2] += p
            p += 1
        count = 0
for i in d:
    if i<=40000 and k//i<=40000:
        ans+=a[i]*b[k//i]+(b[i]*a[k//i])*(i!=k//i)
print(ans)