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 java.util.*;
import java.io.*;
public class Main
{
    public static void main(String args[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        String str[]=br.readLine().split(" ");
        int n=Integer.parseInt(str[0]);
        int m=Integer.parseInt(str[1]);
        int k=Integer.parseInt(str[2]);
        int arr1[]=new int[n];
        int arr2[]=new int[m];
        str=br.readLine().split(" ");
        for(int i=0;i<n;i++)
        arr1[i]=Integer.parseInt(str[i]);
        str=br.readLine().split(" ");
        for(int i=0;i<m;i++)
        arr2[i]=Integer.parseInt(str[i]);
        long consinrow[]=new long[m+1];
        long consincol[]=new long[n+1];
        int cntr=0;
        for(int i=0;i<m;i++)
        {
            if(arr2[i]==0)
            {
                for(int j=1;j<=cntr;j++)
                consinrow[j]+=(cntr-j+1);
                cntr=0;
            }
            else
            cntr++;
        }
        if(cntr!=0)
        {
            for(int j=1;j<=cntr;j++)
            consinrow[j]+=(cntr-j+1);
        }
        cntr=0;
        for(int i=0;i<n;i++)
        {
            if(arr1[i]==0)
            {
                for(int j=1;j<=cntr;j++)
                consincol[j]+=(cntr-j+1);
                cntr=0;
            }
            else
            cntr++;
        }
        if(cntr!=0)
        {
            for(int i=1;i<=cntr;i++)
            consincol[i]+=(cntr-i+1);
        }
        long tot=0;
        for(int i=1;i*i<=k;i++)
        {
            if(k%i==0)
            {
                if(i==k/i)
                {
                    if(i<=m&&i<=n)
                    tot+=(consinrow[i]*consincol[i]);
                }
                else
                {
                    if(i<=m&&k/i<=n)
                    tot+=(consinrow[i]*consincol[k/i]);
                    if(i<=n&&k/i<=m)
                    tot+=(consincol[i]*consinrow[k/i]);
                }
            }
        }
        pw.println(tot);
        pw.flush();
        pw.close();
    }
}