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.io.BufferedInputStream;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
import java.util.TreeSet;

public class B {

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in, 1 << 12));
        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();

        TreeSet<Integer> divs = new TreeSet<>();
        for (int i = 1; i <= Math.min(40_000, n*m); i++) {
            if (k % i == 0) {
                divs.add(i);
                divs.add(k/i);
            }
        }

        int runLen = 0;
        ArrayList<Integer> rls = new ArrayList<>();
        HashMap<Integer, Integer> nDivCounts = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int x = in.nextInt();
            if (x == 0) {
                if (runLen > 0) {
                    rls.add(runLen);
                }
                runLen = 0;
            } else {
                ++runLen;
            }
        }
        if (runLen > 0) {
            rls.add(runLen);
        }
        for (Integer rl : rls) {
            for (Integer div : divs) {
                if (div > rl) {
                    break;
                }
                nDivCounts.put(div, rl - div + 1 + nDivCounts.getOrDefault(div, 0));
            }
        }

        runLen = 0;
        rls = new ArrayList<>();
        HashMap<Integer, Integer> mDivCounts = new HashMap<>();
        for (int i = 0; i < m; i++) {
            int x = in.nextInt();
            if (x == 0) {
                if (runLen > 0) {
                    rls.add(runLen);
                }
                runLen = 0;
            } else {
                ++runLen;
            }
        }
        if (runLen > 0) {
            rls.add(runLen);
        }
        for (Integer rl : rls) {
            for (Integer div : divs) {
                if (div > rl) {
                    break;
                }
                mDivCounts.put(div, rl - div + 1 + mDivCounts.getOrDefault(div, 0));
            }
        }

        long result = 0;
        for (Integer div : divs) {
            result += mDivCounts.getOrDefault(div, 0) * nDivCounts.getOrDefault(k/div, 0);
        }
        System.out.println(result);
    }
}
