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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Solution{
	
	static long mod = 998244353L;
	
	public static void main(String[] args) throws Exception{

		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		
		
		int n = fs.nextInt(), m = fs.nextInt(), k = fs.nextInt();
		int[] a = fs.readArray(n);
		int[] b = fs.readArray(m);
		
		int[] arr1 = func(a);
		int[] arr2 = func(b);
		
		long ans =0;
		
		for(int i=1;i<=n;i++) {
			if(k%i==0 && k/i<=m) {
				ans += (long)arr1[i]*arr2[k/i];
			}
		}
		
		out.println(ans);
		
		
		
		out.close();
		
	}
	
	static int[] func(int[] a) {
		int n = a.length;
		
		int[] count = new int[n+1];
		
		int i=0;
		while(i<n) {
			int d = 0;
			while(i<n && a[i]==1) {
				d++;
				i++;
			}
			
			if(d>0) {
				for(int j=1;j<=d;j++) {
					count[j] += d-j+1;
				}
			}
			i++;
		}
		return count;
	}
	
	

	
	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 String nextLine() throws IOException {
			return br.readLine();
		}
		
		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());
		}
	}
	
}
