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 HelloWorld{
    
    private static int binarySearch(int arr[], int value){
        int low = 0, high = arr.length-1;
        
        while(low <= high){
            int mid = (low + high)>>1;
            
            if(arr[mid] < value)
                low = mid + 1;
            else
                high = mid - 1;
        }
        
        return low;
    }

    public static void main(String []args)throws IOException{
    
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        
        StringTokenizer tk = new StringTokenizer(reader.readLine());
        
        int n = Integer.parseInt(tk.nextToken()),
            m = Integer.parseInt(tk.nextToken()),
            k = Integer.parseInt(tk.nextToken());
        
        int a[] = new int[n], b[] = new int[m];
        
        tk = new StringTokenizer(reader.readLine());
        for(int i=0; i<n; i++)
            a[i] = Integer.parseInt(tk.nextToken());
            
        tk = new StringTokenizer(reader.readLine());
        for(int i=0; i<m; i++)
            b[i] = Integer.parseInt(tk.nextToken());
            
        for(int i=1; i<n; i++){
            if(a[i] == 0)   continue;
            else    a[i] += a[i-1];
        }
        
        for(int i=1; i<m; i++){
            if(b[i] == 0)   continue;
            else    b[i] += b[i-1];
        }
        
        Arrays.sort(a);
        Arrays.sort(b);
        
        long ans = 0;
        
        for(int i=1; i<=Math.sqrt(k); i++){
            if(k%i == 0){
                int j = k/i;
                
                ans += (n-binarySearch(a, i))*(m-binarySearch(b, j));
                if(i != j)
                    ans += (n-binarySearch(a, j))*(m-binarySearch(b, i));
            }
        }
        
        System.out.println(+ans);
    }
}