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 Solution{
	long rem = 1000000007L;
	public Solution(){
		FastReader sc = new FastReader();
		int tests = 1;
        
        for(int t=1; t<=tests; t++){
			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();
			}
			
			int[] adp = new int[n+1];
			int[] bdp = new int[m+1];
			
			int sum = 0;
			for(int i=0; i<n; i++){
				if(a[i] == 0){
					for(int j=1; j<=sum; j++){
						adp[j] += (sum-j+1);
					}
					sum = 0;
				}
				else{
					sum++;
				}
			}
			for(int j=1; j<=sum; j++){
				adp[j] += (sum-j+1);
			}
			
			sum = 0;
			for(int i=0; i<m; i++){
				if(b[i] == 0){
					for(int j=1; j<=sum; j++){
						bdp[j] += (sum-j+1);
					}
					sum = 0;
				}
				else{
					sum++;
				}
			}
			for(int j=1; j<=sum; j++){
				bdp[j] += (sum-j+1);
			}
			sum = 0;
			
			long res = 0L;
			
			for(int i=n; i>0; i--){
				if(k%i > 0){
					continue;
				}
				
				int j = k/i;
				if(j > m){
					break;
				}
				
				res += (adp[i]*(bdp[j]*1L));
			}
			
			System.out.println(res);
        }
	}
	
	public static void main(String[] args){
		new Solution();
	}
}

class FastReader 
{ 
	BufferedReader br; 
	StringTokenizer st; 

	public FastReader() 
	{ 
		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()); 
	} 

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