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.*;
import java.util.StringTokenizer;

public class tank {

    static final FastScanner fs = new FastScanner();

    //static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int n = fs.nextInt(), m = fs.nextInt(), k = fs.nextInt();
        int[] a = fs.readArray(n);
        int[] b = fs.readArray(m);
        int[] len_a = new int[n + 1];
        int[] len_b = new int[m + 1];

        fn(len_a, a, k);
        fn(len_b, b, k);

        long ans = 0;

        for (int i = 1; i <= k && i <= n; i++) {
            if(k % i == 0 && k/i <= m) {
                int q = k/i;

                ans += (long)len_a[i] * len_b[q];
            }
        }

        System.out.println(ans);
    }

    static void fn(int[] len_a, int[] a, int k) {
        int cur = 0, n = a.length;

        for(int i = 0;i < n;i++) {
            if(a[i] == 1) cur++;
            else{
                for (int j = 1; j <= cur && j <= k; j++) {
                    len_a[j] += cur - j + 1;
                }

                cur = 0;
            }

            if(i == n - 1 && a[i] == 1) {
                for (int j = 1; j <= cur && j <= k; j++) {
                    len_a[j] += cur - j + 1;
                }

                cur = 0;
            }
        }
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        String nextLine(){
            try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return "";
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}
