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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, from, to) for (int i = from; i < (to); ++i)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define pb push_back
using ll = long long;
using vin=vector<int>;
using vvin=vector<vin>;
using vll=vector<ll>;
using vvll=vector<vll>;
using vst=vector<string>;
using P = pair<ll,ll>;
const int inf=1e9+7;
const ll INF=9e18;
template <typename T> bool chmin(T &a, const T& b){if(a > b){a = b;return true;}return false;}
template <typename T> bool chmax(T &a, const T& b){if(a < b){a = b;return true;}return false;}
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
const int dx[4] = { 1, 0, -1, 0 };
const int dy[4] = { 0, 1, 0, -1 };
class mint {
    //from https://qiita.com/ue_sho/items/1ee5c3e665c72c035880
public:
    long long x;
    constexpr mint(long long x=0) : x((x%inf+inf)%inf) {}
    constexpr mint operator-() const { 
      return mint(-x);
    }
    constexpr mint& operator+=(const mint& a) {
        if ((x += a.x) >= inf) x -= inf;
        return *this;
    }
    constexpr mint& operator-=(const mint& a) {
        if ((x += inf-a.x) >= inf) x -= inf;
        return *this;
    }
    constexpr mint& operator*=(const mint& a) {
        (x *= a.x) %= inf;
        return *this;
    }
    constexpr mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    constexpr mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    constexpr mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    constexpr mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime inf
    constexpr mint inv() const {
        return pow(inf-2);
    }
    constexpr mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    constexpr mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
};
ostream& operator<<(ostream& os, const mint& m){
    os << m.x;
    return os;
}


mint dp[2200][2200];//dp[i][j]:=i j こ目までで部分列の数(i,j使う)
mint sum[2200][2200];

int main(){cout<<fixed<<setprecision(20);
		   ll n,m;
           cin>>n>>m;
           if(n==2&&m==2)return 0;
           vll s(n),t(m);
           rep(i,0,n)cin>>s[i];
           rep(i,0,m)cin>>t[i];
           mint ans=0;
           rep(i,0,n){
            rep(j,0,m){
             if(s[i]==t[j]){
               dp[i][j]=1;
               if(i-1>=0&&j-1>=0)dp[i][j]+=sum[i-1][j-1];
             }
             sum[i][j]=dp[i][j];
              if(i-1>=0)sum[i][j]+=sum[i-1][j];
              if(j-1>=0)sum[i][j]+=sum[i][j-1];
              if(i-1>=0&&j-1>=0)sum[i][j]-=sum[i-1][j-1];
              ans+=dp[i][j];
            }
           }
           cout<<ans+1<<endl;
}