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 math
n,m,k=[int(x) for x in input().split()]
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a_one=[]
b_one=[]
i=t=0

while i < n:
    if a[i] == 1:
        t+=1
    else:
        a_one.append(t)
        t=0
    i+=1
a_one.append(t)
i=t=0
while i < m:
    if b[i] == 1:
        t+=1
    else:
        b_one.append(t)
        t=0
    i+=1
b_one.append(t)

def solve(r,c,a_one,b_one):
    p=q=i=0
    while i < len(a_one):
        p+=max(a_one[i]-(r-1),0)
        i+=1
    i=0
    while i<len(b_one):
        q+=max(b_one[i]-(c-1),0)
        i+=1
    return (p*q)
ans=0
for i in range(1,(int(math.sqrt(k))+1)):
    if k%i == 0:
        c=k//i
        #print(c)
        ans+= solve(i,c,a_one,b_one)
        if i != c:
            ans+= solve(c,i,a_one,b_one)
print(ans)


