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: Wrong Answer)

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;

public class Main {
	static FastReader scan;
	static PrintWriter out;

	public static void main(String[] args) throws FileNotFoundException {
		Solver solver = new Solver();
		scan = new FastReader();
		out = new PrintWriter(System.out);
		int testCases = 1;
		for(int i = 1; i <= testCases; i++) {
//			out.print("Case #" + i + ": ");
			solver.solve();
		}
		out.close();
	}

	static class Solver {
		
		void solve() {
			int a = scan.nextInt();  
	        int b = scan.nextInt();
	        int mod = 1_000_000_007;
	        int[] aa = scan.nextIntArray(a), bb = scan.nextIntArray(b);
	        long dp[][] = new long[a+1][b+1];  
	        for (int i = 0; i <= a; i++) {  
	            for (int j = 0; j <= b; j++) {  
	                dp[i][j] = 0;  
	            }  
	        }  
	        for (int i = 1; i <= a; i++) {  
	            for (int j = 1; j <= b; j++) {     
	                if (aa[i-1] == bb[j-1])   
	                    dp[i][j] = (1%mod + dp[i][j - 1]%mod + dp[i - 1][j]%mod)%mod;              
	                else 
	                    dp[i][j] = (dp[i][j - 1]%mod + dp[i - 1][j]%mod -   
	                                            dp[i - 1][j - 1]%mod)%mod;              
	            }  
	        }
	        dp[a][b] = (dp[a][b]+1)%mod;
	        out.println(dp[a][b]);
		}
	}

	// Sathvik's Template Stuff BELOW!!!!!!!!!!!!!!!!!!!!!!

	static class DSU {
		int[] root, size;
		int n;

		DSU(int n) {
			this.n = n;
			root = new int[n];
			size = new int[n];
			for (int i = 0; i < n; i++) {
				root[i] = i;
				size[i] = 1;
			}
		}

		int findParent(int idx) {
			while (root[idx] != idx) {
				root[idx] = root[root[idx]];
				idx = root[idx];
			}
			return idx;
		}

		boolean union(int x, int y) {
			int parX = findParent(x);
			int parY = findParent(y);
			if (parX == parY)
				return false;
			if (size[parX] < size[parY]) {
				root[parY] = parX;
				size[parX] += size[parY];
			} else {
				root[parX] = parY;
				size[parY] += size[parX];
			}
			return true;
		}
	}

	static class Extra {
		static void sort(int[] a) {
			Integer[] aa = new Integer[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static void sort(long[] a) {
			Long[] aa = new Long[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static void sort(double[] a) {
			Double[] aa = new Double[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static void sort(char[] a) {
			Character[] aa = new Character[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static long gcd(long a, long b) {
			while (b > 0) {
				long temp = b;
				b = a % b;
				a = temp;
			}
			return a;
		}

		static long lcm(long a, long b) {
			return a * (b / gcd(a, b));
		}

		static boolean isPrime(long n) {
			if (n <= 1)
				return false;
			if (n <= 3)
				return true;
			if (n % 2 == 0 || n % 3 == 0)
				return false;
			for (long i = 5; i * i <= n; i = i + 6) {
				if (n % i == 0 || n % (i + 2) == 0)
					return false;
			}
			return true;
		}

		static HashSet<Integer> sieve(int n) {
			boolean[] prime = new boolean[n + 1];
			HashSet<Integer> res = new HashSet<>();
			for (int p = 2; p * p <= n; p++) {
				if (!prime[p]) {
					res.add(p);
					for (int i = p * p; i <= n; i += p)
						prime[i] = true;
				}
			}
			return res;
		}

		static HashMap<Long, Integer> primeFactorization(long n) {
			HashMap<Long, Integer> res = new HashMap<>();
			while (n % 2 == 0) {
				res.put(2L, res.getOrDefault(2L, 0) + 1);
				n /= 2;
			}
			for (long i = 3; i * i <= n; i += 2) {
				while (n % i == 0) {
					res.put(i, res.getOrDefault(i, 0) + 1);
					n /= i;
				}
			}
			if (n > 2)
				res.put(n, 1);
			return res;
		}
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		public FastReader(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

		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());
		}

		int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		long[] nextLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		double[] nextDoubleArray(int n) {
			double[] a = new double[n];
			for (int i = 0; i < n; i++)
				a[i] = nextDouble();
			return a;
		}

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

}