Problem:
Score : 500 points

Problem Statement

You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).

In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?

Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.

For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.

Since the answer can be tremendous, print the number modulo 10^9+7.

Constraints

1 \leq N, M \leq 2 \times 10^3

The length of S is N.

The length of T is M.

1 \leq S_i, T_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}

Output

Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.

Sample Input 1

2 2
1 3
3 1

Sample Output 1

3

S has four subsequences: (), (1), (3), (1, 3).

T has four subsequences: (), (3), (1), (3, 1).

There are 1 \times 1 pair of subsequences in which the subsequences are both (), 1 \times 1 pair of subsequences in which the subsequences are both (1), and 1 \times 1 pair of subsequences in which the subsequences are both (3), for a total of three pairs.

Sample Input 2

2 2
1 1
1 1

Sample Output 2

6

S has four subsequences: (), (1), (1), (1, 1).

T has four subsequences: (), (1), (1), (1, 1).

There are 1 \times 1 pair of subsequences in which the subsequences are both (), 2 \times 2 pairs of subsequences in which the subsequences are both (1), and 1 \times 1 pair of subsequences in which the subsequences are both (1,1), for a total of six pairs.
Note again that we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.

Sample Input 3

4 4
3 4 5 6
3 4 5 6

Sample Output 3

16

Sample Input 4

10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7

Sample Output 4

191

Sample Input 5

20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Sample Output 5

846527861

Be sure to print the number modulo 10^9+7.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main{
    static long mod=1000000007;
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};

    public static void main(String[] args)   throws Exception, IOException{        
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

        int n = sc.nextInt();
        int m = sc.nextInt();
        int a[]=sc.nextIntArray(n);
        int b[]=sc.nextIntArray(m);
        long dp[][] = new long[n+1][m+1];
        fill(dp[0],1);
        for (int i = 1; i <= n; i++) {
        	dp[i][0]=1;
        }
        
        for (int i = 1; i <= n; i++) {
			for (int t = 1; t <= m ; t++) {
				dp[i][t] = dp[i][t-1] + dp[i-1][t]%mod - dp[i-1][t-1];
				dp[i][t]%=mod;
				if(a[i-1]==b[t-1])dp[i][t]+=dp[i-1][t-1];
				dp[i][t]%=mod;
				if(dp[i][t]<0)dp[i][t]+=mod;
//				db(i,t,dp[i][t]);
			}
		}
        
        out.println(dp[n][m]);
        out.flush();
    }

    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}
//class P {
//    int id, d;
//    P(int  id, int d) {
//        this.id=id;
//        this.d=d;
//    }
//}

//class P implements Comparable<P>{
//    int a,b;
//    double c;
//    P(int a, int b) {
//    	this.b=b;
//    	this.a=a;
//    	this.c=(double)b/a;
//    }
//
//    public int compareTo(P p){
//        return Double.compare(p.c,c);
//    }
//}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) throws IOException{
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) throws IOException{
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}