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 {
  public static void update(int ct[], int val)
  {
    for(int i=1;i<val+1;i++)
    {
      ct[i]+=(val-i+1);
    }
  }
  public static void main(String[] args)throws IOException {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        int a[] = new int[n];
        int cta[] = new int[n+1];
        int b[] = new int[m];
        int ctb[] = new int[m+1];
        for(int i=0;i<n;i++) {a[i]=in.nextInt();}
        for(int i=0;i<n+1;i++)
          cta[i]=0;
        for(int i=0;i<m+1;i++)
          ctb[i]=0;
        for(int i=0;i<m;i++) b[i]=in.nextInt();
        int ct=0;
        for(int i=0;i<n;i++)
        {
          if(a[i]==0)
          {
            update(cta,ct);
            ct=0;
          }
          else
            ct++;
        }
        update(cta,ct);
        ct=0;
        for(int i=0;i<m;i++)
        {
          if(b[i]==0)
          {
            update(ctb,ct);
            ct=0;
          }
          else
            ct++;
        }
        update(ctb,ct);
        long ans = 0;
        // for(int i=0;i<n+1;i++)
        //   out.println("cta i "+i+"val "+cta[i]);
        // for(int i=0;i<m+1;i++)
        //   out.println("ctb i "+i+"val "+ctb[i]);
        for(int i=1;i<=(int)Math.sqrt(k);i++)
        {
          if(k%i!=0)
            continue;
          int j = k/i;
          // out.println("i="+i+",j="+j);
          if(i<n+1 && j<m+1){
            // out.println("yo1");
            ans+=(cta[i]*ctb[j]);
          }
          if((i!=j) && j<n+1 && i<m+1){
            // out.println("yo2");
            ans+=(cta[j]*ctb[i]);
          }
        }
        out.println(ans);
        out.close();
      }
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
 
    }
 }