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 Main {
    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int k = scanner.nextInt();

        HashMap<Integer,Integer> k1 = new HashMap<>();
        HashMap<Integer,Integer> k2 = new HashMap<>();

        LinkedList<Integer> divisors = new LinkedList<>();
        int s = 1;
        while (s <= Math.sqrt(k)){
            if (k % s == 0){
                k1.put(s,0);
                k1.put(k/s,0);
                k2.put(s,0);
                k2.put(k/s,0);
                divisors.add(s);
                if (s != k/s)
                    divisors.add(k/s);
            }
            s++;
        }

        Collections.sort(divisors);

        int size = divisors.size();
        TreeMap<Integer,Integer> treeMapk1 = new TreeMap<>();
        treeMapk1.putAll(k1);
        TreeMap<Integer,Integer> treeMapk2 = new TreeMap<>();
        treeMapk2.putAll(k2);

        int t = 0;
        for (int i = 0; i < n; i++) {
            if (scanner.nextInt() == 1)
                t++;
            else{
                int j = 0;
                while (j < size && divisors.get(j) <= t){
                    int u = k1.get(divisors.get(j));
                    k1.replace(divisors.get(j),u+t-divisors.get(j)+1);
                    j++;
                }

                t = 0;
            }
        }

        int j = 0;
        while (j < size && divisors.get(j) <= t) {
            int u = k1.get(divisors.get(j));
            k1.replace(divisors.get(j), u + t - divisors.get(j) + 1);
            j++;
        }


        t = 0;
        for (int i = 0; i < m; i++) {
            if (scanner.nextInt() == 1)
                t++;
            else {
                j = 0;
                while (j < size && divisors.get(j) <= t){
                    int u = k2.get(divisors.get(j));
                    k2.replace(divisors.get(j),u+t-divisors.get(j)+1);
                    j++;
                }
                t = 0;
            }
        }

        j = 0;
        while (j < size && divisors.get(j) <= t) {
            int u = k2.get(divisors.get(j));
            k2.replace(divisors.get(j), u + t - divisors.get(j) + 1);
            j++;
        }


        long sum = 0;
        for (int i = 0; i < size; i++) {
            sum += (k1.get(divisors.get(i)))*k2.get(k/divisors.get(i));
        }
        System.out.println(sum);

    }
}
