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 A {

    static int[] solve(int arr[], int n) {
        int ans[] = new int[n + 1];

        for (int i = 0; i < n; i++) {
            if (arr[i] == 0) {
                continue;
            }

            int j = i;
            while (j < n && arr[j] == 1) {
                j++;
            }

            j--;

            int len = j - i + 1;

            for (int k = len; k >= 1; k--) {
                ans[1] = ans[1] + 1;

                if (k + 1 <= n) {
                    ans[k + 1] = ans[k + 1] - 1;
                }

            }

            i = j;

        }

        for (int i = 1; i < n + 1; i++) {
            ans[i] = ans[i] + ans[i - 1];
        }

        return ans;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();

        int n = sc.nextInt();
        int m = sc.nextInt();

        int k = sc.nextInt();

        int a[] = new int[n];
        int b[] = new int[m];

        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        for (int i = 0; i < m; i++) {
            b[i] = sc.nextInt();
        }

        ArrayList<Integer> factors = new ArrayList<>();

        int sqrt = (int) Math.sqrt(k);

        for (int i = 1; i <= sqrt; i++) {
            if (k % i != 0) {
                continue;
            }

            if (i == k / i) {
                factors.add(i);
            } else {
                factors.add(i);
                factors.add(k / i);
            }

        }

        int pre_a[] = solve(a, n);
        int pre_b[] = solve(b, m);

        long ans = 0;

        // for (int i = 0; i <= n; i++) {
        // System.out.print(pre_a[i] + " ");
        // }

        // System.out.println();

        // for (int i = 0; i <= m; i++) {
        // System.out.print(pre_b[i] + " ");
        // }

        // System.out.println();

        for (int i = 0; i < factors.size(); i++) {
            int x = factors.get(i);
            int y = k / x;

            // System.out.println(x + " " + y);

            if (x <= n && y <= m) {
                ans = ans + pre_a[x] * pre_b[y];
            }

        }

        sb.append(ans).append("\n");

        System.out.print(sb);
    }
}