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.*;
public class p1323B {
    public static void main(String[] args) {
        Scanner scn=new Scanner(System.in);
        int n=scn.nextInt();
        int m=scn.nextInt();
        int k=scn.nextInt();
        int[] a=new int[n];
        int[] b=new int[m];
        for(int i=0;i<n;i++){
            a[i]=scn.nextInt();
        }
        for(int i=0;i<m;i++){
            b[i]=scn.nextInt();
        }
        HashMap<Integer,Integer> a1=new HashMap<Integer,Integer>();
        int j=0;
        while(j<n){
            int cnt=0;
            while(j+cnt<n&&a[j]==a[j+cnt]){
                cnt++;
            }
            if(a[j]==1){
                for(int i=1;i<=cnt;i++){
                    if(a1.keySet().contains(i)==false){
                        a1.put(i,(cnt-i+1));
                    }
                    else{
                        a1.put(i,a1.get(i)+(cnt-i+1));
                    }
            }
            }
            j=j+cnt;
        }
        HashMap<Integer,Integer> b1=new HashMap<>();
        j=0;
        while(j<m){
            int cnt=0;
            while(j+cnt<m&&b[j]==b[j+cnt]){
                cnt++;
            }
            if(b[j]==1){
                for(int i=1;i<=cnt;i++){
                    if(b1.keySet().contains(i)==false){
                        b1.put(i,(cnt-i+1));
                    }
                    else{
                        b1.put(i,b1.get(i)+(cnt-i+1));
                    }
            }
            }
            j=j+cnt;
        }
        long ans=0;
        for(int i=1;i*i<=k;i++){
            if(k%i==0){
                if(b1.get(k/i)!=null&&a1.get(i)!=null){
                    ans=ans+(long)(b1.get(k/i))*(long)(a1.get(i));
                } 
            }
        }
        for(int i=1;i*i<k;i++){
            if(k%i==0){
                if(b1.get(i)!=null&&a1.get(k/i)!=null){
                    ans=ans+(long)(b1.get(i))*(long)(a1.get(k/i));
                }   
            }
        }
        System.out.println(ans);
    }
}
