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)

/**
 * Created at 20:53 on 2019-06-16
 */

import java.io.*;
import java.util.*;

public class Main {

  static FastScanner sc = new FastScanner();

  static int[] dx = {0, 1, 0, -1};
  static int[] dy = {-1, 0, 1, 0};

  static long MOD = (long) (1e9 + 7);

  public static void main(String[] args) {

    int N = sc.nextInt();
    int M = sc.nextInt();

    int[] S = sc.nextIntArray(N);
    int[] T = sc.nextIntArray(M);

    long[][] dp = new long[N+1][M+1];
    /*
    for (int i=0; i<=N; i++) {
      dp[i][0] = 1;
    }
    for (int j=0; j<=M; j++) {
      dp[0][j] = 1;
    }
    */
    for (int i=1; i<=N; i++) {
      for (int j=1; j<=M; j++) {
        dp[i][j] = (dp[i][j-1] + dp[i-1][j] - dp[i-1][j-1]) % MOD;
        if(S[i-1] == T[j-1]) {
          dp[i][j] = (dp[i][j] + dp[i-1][j-1] + 1) % MOD;
        }
      }
    }

    System.out.println(dp[N][M]+1);

  }

  public static class Mathf {

    public static int max(int[] a) {
      int M = a[0];
      for (int i = 1; i < a.length; i++) {
        M = Math.max(M, a[i]);
      }
      return M;
    }

    public static int min(int[] a) {
      int m = a[0];
      for (int i = 1; i < a.length; i++) {
        m = Math.min(m, a[i]);
      }
      return m;
    }

    public static long max(long[] a) {
      long M = a[0];
      for (int i = 1; i < a.length; i++) {
        M = Math.max(M, a[i]);
      }
      return M;
    }

    public static long min(long[] a) {
      long m = a[0];
      for (int i = 1; i < a.length; i++) {
        m = Math.min(m, a[i]);
      }
      return m;
    }

  }

  /*
    add()でインデックスを指定しない場合指定されたソート順に違わない位置に追加する
    (ただしソートされていることが前提となる)
    Comparatorが0を返したとき、それらの順序は保証しない
    (TreeSet, TreeMapと違い削除はしない)
   */
  static class TreeList<E> extends ArrayList<E> {

    Comparator<? super E> comparator;

    TreeList(Comparator<? super E> c) {
      super();
      comparator = c;
    }

    /*
      ソート済みのリストに要素を追加する
     */
    public boolean add(E e) {
      int lowIndex = 0;
      int highIndex = size() - 1;
      int index = 0;

      if (size() == 0) {
        super.add(e);
        return true;
      }

      if (comparator.compare(e, get(0)) < 0) {
        index = 0;
      } else if (comparator.compare(e, get(highIndex)) > 0) {
        index = highIndex + 1;
      } else {
        while (lowIndex <= highIndex) {

          if (highIndex == lowIndex + 1 || highIndex == lowIndex) {
            index = highIndex;
            break;
          }

          int midIndex = (lowIndex + highIndex) / 2;
          ;

          if (comparator.compare(e, get(midIndex)) > 0) {
            lowIndex = midIndex;
          } else {
            highIndex = midIndex;
          }

        }
      }

      super.add(index, e);
      return true;
    }

  }

  static class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = in.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) return buffer[ptr++];
      else return -1;
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
      skipUnprintable();
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) throw new NoSuchElementException();
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) throw new NoSuchElementException();
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }

    public int nextInt() {
      return (int) nextLong();
    }

    public int[] nextIntArray(int N) {
      int[] array = new int[N];
      for (int i = 0; i < N; i++) {
        array[i] = sc.nextInt();
      }
      return array;
    }

    public long[] nextLongArray(int N) {
      long[] array = new long[N];
      for (int i = 0; i < N; i++) {
        array[i] = sc.nextLong();
      }
      return array;
    }
  }

}