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 collections import Counter as C
n,m,k=map(int,input().split())
a=list(map(int,input().split()))+[0]
b=list(map(int,input().split()))+[0]
c,d=[],[]
for i in range(1,n+1):
    if a[i]==0:
        c.append(a[i-1])
    else:
        a[i]=a[i-1]+1
for i in range(1,m+1):
    if b[i]==0:
        d.append(b[i-1])
    else:
        b[i]=b[i-1]+1
a=sorted(C(c).items(),key=lambda x:x[0])
b=sorted(C(d).items(),key=lambda x:x[0])
s,t=len(a),len(b)
ans=0
for i in range(1,int(k**0.5)+1):
    if k%i==0:
        c,d=0,0
        for j in range(s-1,-1,-1):
            if a[j][0]>=i:
                c+=(a[j][0]-i+1)*a[j][1]
            if a[j][0]>=k//i:
                d+=(a[j][0]-k//i+1)*a[j][1]
            elif a[j][0]<i and a[j][0]<k//i:
                break
        e,f=0,0
        for j in range(t-1,-1,-1):
            if b[j][0]>=i:
                e+=(b[j][0]-i+1)*b[j][1]
            if b[j][0]>=k//i:
                f+=(b[j][0]-k//i+1)*b[j][1]
            elif b[j][0]<i and b[j][0]<k//i:
                break
        if i!=k//i:
            ans+=c*f+e*d
        else:
            ans+=c*f
print(ans)