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
l=input().split()
n=int(l[0])
m=int(l[1])
k=int(l[2])
l=[]
i=1
a1=input().split()
ai=[int(i) for i in a1]
b1=input().split()
bi=[int(i) for i in b1]
while(i*i<=k):
    if(k%i==0 and i!=k//i):
        l.append((i,k//i))
        l.append((k//i,i))
    elif(k%i==0):
        l.append((i,i))
    i+=1

arr1=[0 for i in range(n)]
arr2=[0 for i in range(m)]
count1=0
for i in ai:
    if(i==1):
        count1+=1
    else:
        for j in range(count1):
            arr1[j]+=count1-j
        count1=0

for i in range(count1):
    arr1[i]+=count1-i
count1=0
for i in bi:
    if(i==1):
        count1+=1
    else:
        for j in range(count1):
            arr2[j]+=count1-j
        count1=0
for i in range(count1):
    arr2[i]+=count1-i
ans=0

for i in l:
    if(i[0]>n or i[1]>m):
        ans=ans
    else:
        ans+=(arr1[i[0]-1]*arr2[i[1]-1])

print(ans)
