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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
using namespace std;

#define rep(i, s) for (int i = 0; i < s; ++i)
#define all(v) (v.begin(), v.end())
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define VEC(a, n) vector<int>a(n)
#define PQ(a) priority_queue<int>a
#define PQmin(a) priority_queue< int, :vector<int>, greater<int> >a
#define PAIR pair<int, int>
/*
sort(sp.begin(), sp.end(), [](PAIR l, PAIR r){
        return l.first<r.first || (l.first==r.first && l.second > r.second);
        });
*/

typedef long long  ll;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

int main() {
    ll p=pow(10,9)+7;
    int n,m;
    cin>>n>>m;
    vector<int>s(n);
    vector<int>t(m);
    for(int i=0;i<n;i++){
        cin>>s.at(i);
    }
    for(int i=0;i<m;i++){
        cin>>t.at(i);
    }

    //二次元dpの実装
    ll ans[n+1][m+1];
    for(int i=0;i<n+1;i++){
        for(int j=0;j<m+1;j++){
            if(i==0){
                ans[i][j]=1;
            }
            else if(j==0){
                ans[i][j]=1;
            }
            else{
                ans[i][j]=0;
            }
        }
    }
    // for(int i=0;i<n+1;i++){
    //     for(int j=0;j<m+1;j++){
    //         cout<<ans[i][j]<<" ";
    //     }
    //     cout<<endl;
    // }

    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(s.at(i)==t.at(j)){//一緒だった場合、(965,945)みたいなペアだとすると、(965,94)が持っていたペアの末尾に5をつけてもそれがまた条件を満たすペアになるので、2倍。
                                //さらに、(96,945)が持っていたペアに5をつけてもそれはまた条件を満たすので2倍。しかし、(96,94)のペアに5をつけたものがダブっているのでそれをひく。
                ans[i+1][j+1]=ans[i][j+1]+ans[i+1][j];
            }
            else{
                ans[i+1][j+1]=ans[i][j+1]+ans[i+1][j]-ans[i][j];
            }
            ans[i+1][j+1]%=p;
        }
    }
    cout<<ans[n][m]%p<<endl;
    // for(int i=0;i<n+1;i++){
    //     for(int j=0;j<m+1;j++){
    //         cout<<ans[i][j]<<" ";
    //     }
    //     cout<<endl;
    // }





    return 0;
}