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.Scanner;

public class countsubrectangles 
{
	public static void main(String[] args) 
	{
		Scanner inp = new Scanner(System.in);
		int n = inp.nextInt();
		int m = inp.nextInt();
		int k = inp.nextInt();
		int[] a = new int[n];
		int[] b = new int[m];
		for(int i=0;i<n;i++)
			a[i] = inp.nextInt();
		for(int i=0;i<m;i++)
			b[i] = inp.nextInt();
		
		int[] sa = new int[n+1];
		int[] sb = new int[m+1];
		for(int i=0;i<=n;i++)
			sa[i] = 0;
		for(int i=0;i<=m;i++)
			sb[i] = 0;
		
		int flag = 0;
		int size = 0;
		for(int i=0;i<n;i++)
		{
			if(a[i]==1)
			{
				flag = 1;
				size++;
			}
			if((flag==1)&&((i==n-1)||(a[i]==0)))
			{
				for(int j=1;j<=size;j++)
					sa[j] += size - j + 1;
				size = 0;
			}
		}
		
		flag = 0;
		size = 0;
		for(int i=0;i<m;i++)
		{
			if(b[i]==1)
			{
				flag = 1;
				size++;
			}
			if((flag==1)&&((i==m-1)||(b[i]==0)))
			{
				for(int j=1;j<=size;j++)
					sb[j] += size - j + 1;
				size = 0;
			}
		}
	
		long ans = 0;
		for(int i=1;i<=n;i++)             //Loop to go over divisors of k
		{
			if(k%i==0)
			{
				int j = k/i;
				if(j<=m)
				{
					ans += sa[i] * sb[j];
				}
			}
		}
		System.out.println(ans);
	}
}
