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.Scanner;

public class WorkFile {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt(), m = s.nextInt(), k = s.nextInt();
        int[] a = new int[n+1];
        int count = 0;
        for (int i=0; i<n; i++) {
            int x = s.nextInt();
            if (x==1) count++;
            if (x==0 || i==n-1) {
                for (int j=1; j<=count; j++) {
                    a[j]+=count-j+1;
                }
                count=0;
            }
        }
        int[] b = new int[m+1];
        count = 0;
        for (int i=0; i<m; i++) {
            int x = s.nextInt();
            if (x==1) count++;
            if (x==0 || i==m-1) {
                for (int j=1; j<=count; j++) {
                    b[j]+=count-j+1;
                }
                count=0;
            }
        }
        long res = 0;
        for (int i=1; i<=n; i++) {
            if (k%i==0) {
                if (k/i>m) continue;
                res+=a[i]*b[k/i];
            }
        }
        System.out.println(res);
    }
}