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.math.BigInteger;
import java.util.*;
import java.util.Map.Entry;
import java.util.Stack;


public class ROUGH{
	public static class FastReader {
		BufferedReader br;
		StringTokenizer root;
		
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (root == null || !root.hasMoreTokens()) {
				try {
					root = new StringTokenizer(br.readLine());
				} catch (Exception addd) {
					addd.printStackTrace();
				}
			}
			return root.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception addd) {
				addd.printStackTrace();
			}
			return str;
		}
	}
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	
	public static FastReader sc = new FastReader();
	public static void main(String[] args) {
		
			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[] x = fill(a);
			long[] y = fill(b);
			long ans = 0;
			for(int i=1;i<x.length;++i) {
				if(k%i == 0 && k/i <=m) ans+=x[i]*y[k/i];
			}
			out.println(ans);
		
			
		
		
		out.close();
		}
	private static long[] fill(int[] b) {
		int n = b.length;
		long[] freq = new long[n+1];
		for(int i=0;i<n;++i) {
			if(b[i] == 0) continue;
		    int j = i;
		    while(j < n && b[j] == 1) ++j;
		    
		    for(int l=1;l<=j-i;++l) {
		    	freq[l]+=j-i-l+1;
		    }
		    i=j;
		}
		return freq;
	}
	
	}

