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.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        int[] a = new int[n+1];
        int[] b = new int[m+1];
        for(int i = 0; i<n; i++) {
            a[i] = sc.nextInt();
        }
        for(int i = 0; i<m; i++) {
            b[i] = sc.nextInt();
        }
        a[n] = b[m] = 0;
        for(int i = n-1; i>=0; i--) {
            if(a[i] != 0) {
                a[i] = 1 + a[i+1];
            }
        }
        for(int i = m-1; i>=0; i--) {
            if(b[i] != 0) {
                b[i] = 1 + b[i+1];
            }
        }
        Long ans = 0L;
        Map<Integer, Integer> aSubM = new HashMap<>();
        Map<Integer, Integer> bSubM = new HashMap<>();

        for(int p = 0; p<n; ) {
            if(a[p] == 0) {
                p += 1;
                continue;
            }
            for (int i=1; i<=a[p]; i++) {
                aSubM.putIfAbsent(i, 0);
                aSubM.put(i, aSubM.get(i) + a[p] - i + 1);
            }
            p = p + a[p];
        }

        for(int p = 0; p< m; ) {
            if(b[p] == 0) {
                p += 1;
                continue;
            }
            for (int i= 1; i<=b[p]; i++) {
                bSubM.putIfAbsent(i, 0);
                bSubM.put(i, bSubM.get(i) + b[p] - i + 1);
            }
            p = p  + b[p];
        }


        for(int n1 = 1; n1<=n; n1++) {
            if(k % n1 == 0) {
                int n2 = k / n1;
                int aSub = aSubM.get(n1) != null ? aSubM.get(n1) : 0;
                int bSub = bSubM.get(n2) != null ? bSubM.get(n2) : 0;
                ans += aSub*Long.valueOf(bSub);
            }
        }
        System.out.println(ans);
    }

}

