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.*;
import java.io.*;

public class File {
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
	
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
		
		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();
		}
		
		
		
		long[] aCount = new long[n+1];
		long[] bCount = new long[m+1];
		
		int i = 0;
		while (i < n) {
			if (a[i] == 0) {
				i++;
			}
			else {
				int j = i;
				
				while (j + 1 < n && a[j + 1] == 1) {
					j++;
				}
				
				for (int len = 1; len <= j - i + 1; len++) {
					aCount[len] += ((j - i + 1) - len + 1);
				}
				
				i = j + 1;
			}
		}
		i = 0;
		while (i < m) {
			if (b[i] == 0) {
				i++;
			}
			else {
				int j = i;
				
				while (j + 1 < m && b[j + 1] == 1) {
					j++;
				}
				
				for (int len = 1; len <= j - i + 1; len++) {
					bCount[len] += ((j - i + 1) - len + 1);
				}
				
				i = j + 1;
			}
		}
		
		
		
		long total = 0;
		for (int d = 1; d <= Math.sqrt(k); d++) {
			if (k % d == 0) {
				int d2 = k / d;
				
				if (n >= d && m >= d2) {
					total += (aCount[d] * bCount[d2]);
				}
				
				if (d2 != d) {
    				if (n >= d2 && m >= d) {
    				    total += (aCount[d2] * bCount[d]);
    				}
				}
			}
		}
		
		
		out.println(total);
		
		
		
		
		out.close();
	}
}











	