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.awt.*;
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int k=sc.nextInt();
        int a[]=new int[n];
        int b[]=new int[m];
        for (int i=0;i<n;i++)a[i]=sc.nextInt();
        for (int i=0;i<m;i++)b[i]=sc.nextInt();
        int suma[]=new int[n];
        int sumb[]=new int[m];
        suma[0]=a[0];
        sumb[0]=b[0];
        for (int i=1;i<n;i++){
            if (a[i]==1){
                suma[i]=suma[i-1]+1;
            }
        }
        for (int i=1;i<m;i++){
            if (b[i]==1){
                sumb[i]=sumb[i-1]+1;
            }
        }
        int aa[]=new int[n+2];
        int bb[]=new int[m+2];
        for (int i=0;i<n;i++){
            if (suma[i]==0)continue;
            aa[suma[i]+1]--;
            aa[1]++;
        }
        for (int i=0;i<m;i++){
            if (sumb[i]==0)continue;
            bb[sumb[i]+1]--;
            bb[1]++;
        }
        for (int i=1;i<=n;i++){
            aa[i]+=aa[i-1];
        }
        for (int i=1;i<=m;i++){
            bb[i]+=bb[i-1];
        }
        long ans=0;
        for (int i=1;i<=n;i++){
            if (k%i==0 && k/i<=m){
                ans+=aa[i]*bb[k/i];
            }
        }
        System.out.println(ans);
    }
}