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.*;
import java.util.*;
public class CountSubrectangles{
    static Scanner sc = new Scanner(System.in);
    
    static Vector<Integer> findFactors(long p)
    {
        Vector factors = new Vector<Integer>();
        factors.add(1);
        for(int i=2;i<= (int)Math.sqrt(p);i++){
            if(p%i == 0)
                factors.add(i);
        }
        return factors;
    }

    static int findComb(boolean arr[],long side){
        if(side > (long)arr.length)
            return 0;
        int count = 0,out =0;
        Vector v = new Vector<Integer>();

        for(int i=0;i<arr.length;i++){
            if(arr[i])
                count++;
            else if(count > 0)
            {
                if(count >= (int)side)
                    v.add(count);
                count = 0;
            }    
        }
        if(count >= (int)side)
            v.add(count);

        for(int i=0;i<v.size();i++) 
        {
            out = out + (int)v.elementAt(i) - (int)side + 1;
        }   
        v.clear();
        return out;
    }

    public static void main(String[] args) {
        int n,m;
        long p;
        n = sc.nextInt();
        m = sc.nextInt();
        p = sc.nextLong();

        boolean A[] = new boolean[n];
        boolean B[] = new boolean[m];

        for(int i=0;i<n;i++)
        {
            int temp = sc.nextInt();
            if(temp == 0)
                A[i] = false;
            else    
                A[i] = true;    
        }
        for(int i=0;i<m;i++)
        {
            int temp = sc.nextInt();
            if(temp == 0)
                B[i] = false;
            else    
                B[i] = true;    
        }
        
        long count = 0; 
        Vector<Integer> factors = findFactors(p);
        for(int i=0;i< factors.size();i++)
        {
            long sideA = (long)factors.elementAt(i);
            long sideB = p/sideA;
            long temp;
            temp = (long)findComb(A,sideA);
            temp = temp*(long)findComb(B,sideB);
            count+=temp;
            if(sideA != sideB)
            {
                temp = (long)findComb(A,sideB);
                temp = temp*(long)findComb(B,sideA);
                count+=temp;
            }
        }
        System.out.println(count);
    }
}