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>
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000

"""
import sys
input = sys.stdin.buffer.readline

def solution():
    n,m,k=map(int,input().split())
    a=list(map(int,input().split()))
    b=list(map(int,input().split()))
    a.append(0)
    b.append(0)
    d1={}
    d2={}
    l=0
    for i in range(n):
        if a[i]==1:
            l+=1
        if a[i]==0:
            l=0
        if a[i]==1 and a[i+1]==0:
            d1[l]=d1.get(l,0)+1
    l=0
    for i in range(m):
        if b[i]==1:
            l+=1
        if b[i]==0:
            l=0
        if b[i]==1 and b[i+1]==0:
            d2[l]=d2.get(l,0)+1
    ans=0
    for i in range(1,int(k**0.5)+1):
        cnt1=cnt2=cnt3=cnt4=0
        if k%i==0:
            j=k//i
            for z in d1.keys():
                if z>=i:
                    cnt1+=(z-i+1)*(d1[z])
                if z>=j and i!=j:
                    cnt2+=(z-j+1)*(d1[z])
            for z in d2.keys():
                if z>=j:
                    cnt3+=(z-j+1)*(d2[z])
                if z>=i and i!=j:
                    cnt4+=(z-i+1)*(d2[z])
            ans+=(cnt1*cnt3)+(cnt2*cnt4)           
    print(ans)
solution()