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.util.*;
public class Main
{
	
	private void solve()throws IOException
	{
		int n=nextInt();
		int m=nextInt();
		int k=nextInt();

		int trail=0;
		long t[]=new long[n+1];
		for(int i=1;i<=n;i++)
		{
			if(nextInt()==1)
				trail++;
			else
			{
				t[trail]++;
				trail=0;
			}
		}
		t[trail]++;
		trail=0;

		int c[]=new int[n+1];
		for(int i=n;i>=1;i--)
		{
			c[i]=(i==n?0:c[i+1])+(int)t[i];
			t[i]=(i==n?0:t[i+1])+i*t[i];
		}

		long tb[]=new long[m+1];
		for(int i=1;i<=m;i++)
		{
			if(nextInt()==1)
				trail++;
			else
			{
				tb[trail]++;
				trail=0;
			}
		}
		tb[trail]++;
		trail=0;

		int cb[]=new int[m+1];
		for(int i=m;i>=1;i--)
		{
			cb[i]=(i==m?0:cb[i+1])+(int)tb[i];
			tb[i]=(i==m?0:tb[i+1])+i*tb[i];
		}
		
		long ans=0;
		for(int i=1;i*i<=k;i++)
		{
			if(k%i==0)
			{
				if(i<=n && (k/i)<=m)
				{
					long counta=t[i]-c[i]*i+c[i];
					long countb=tb[k/i]-cb[k/i]*(k/i)+cb[k/i];
					ans+=counta*countb;
				}
				if(i*i!=k && i<=m && (k/i)<=n)
				{
					long counta=t[k/i]-c[k/i]*(k/i)+c[k/i];
					long countb=tb[i]-cb[i]*(i)+cb[i];
					ans+=counta*countb;
				}
			}
		}
		out.println(ans);
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws IOException
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		solve();

		br.close();
		out.close();
	}
	public static void main(String args[])throws IOException{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws IOException{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws IOException{
		return br.readLine();
	}
	int nextInt()throws IOException{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws IOException{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws IOException{
		return Double.parseDouble(nextToken());
	}
}