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 sys
input=sys.stdin.readline
import math
n,m,k=map(int,input().split())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
max_hor=[0]*(40005)
max_vert=[0]*(40005)
def subrect_finder(l,l2,nm):
    count=l[0]
    l2[count]=count
    for i in range(1,nm):
        if l[i-1]==l[i] and l[i]==1:
            count+=1
            l2[count]+=1
        elif l[i]==1:
            count+=1
            l2[count]+=1
        else:
            count=0
    return(l2)
hor=(subrect_finder(a,max_hor,n))
vert=(subrect_finder(b,max_vert,m))
ans=0
#print(l1,hor,vert)
for j in  range(1,math.floor(k**0.5)+1):
    if k%j==0:
        jk=int(k/j)
        j,jk=min(40001,j),min(40001,jk)
        #print(jk)
        if vert[j]!=0:
            ans+=sum(vert[j:])*(sum(hor[jk:]))
        if hor[j]!=0 and j!=jk:
            ans+=sum(hor[j:])*(sum(vert[jk:]))
        #print(ans)
print(ans)

    

