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>
//package CfProblemSet.CF626;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class B {
    static int a[], b[];
    static int n, m, k;

    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        n = fs.nextInt();
        m = fs.nextInt();
        k = fs.nextInt();
        a = fs.readArray( n );
        b = fs.readArray( m );
        long ans = 0;
        for (int i = 1; i * i <= k; i++) {
            if (k % i == 0) {
                ans += (i * i == k) ? find( i, i ) : (find( i, k / i ) + find( k / i, i ));
            }
        }
        System.out.println( ans );
    }

    private static long find(int x, int y) {
        long rc = 0, cc = 0;
        long cnt = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] == 1) {
                cnt++;
            } else {
                cnt = 0;
            }
            if (cnt >= x) rc++;
        }
        cnt = 0;
        for (int i = 0; i < m; i++) {
            if (b[i] == 1) {
                cnt++;
            } else {
                cnt = 0;
            }
            if (cnt >= y) cc++;
        }
        return rc * cc;
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader( new InputStreamReader( System.in ) );
        StringTokenizer st = new StringTokenizer( "" );

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer( br.readLine() );
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

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

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