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)

// template version 1.12
using namespace std;
#include <iostream>
#include <bits/stdc++.h>

// varibable settings
#define infile "../test/sample-1.in"
#define int long long //{{{
const int INF=1e18;
const int MOD=1e9+7; //}}}

// define basic macro {{{
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) rrepi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)((b)-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)
#define each(i,a) for (auto&& i : a)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define ub upper_bound
#define lb lower_bound
#define posl(A, x) (lower_bound(all(A), x)-A.begin())
#define posu(A, x) (upper_bound(all(A),x)-A.begin())
template<class T> inline void chmax(T &a, const T &b) { if((a) < (b)) (a) = (b); }
template<class T> inline void chmin(T &a, const T &b) { if((a) > (b)) (a) = (b); }

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;

template<typename T> using PQ = priority_queue<T, vector<T>, greater<T>>;
struct Fast { Fast(){ std::cin.tie(0); ios::sync_with_stdio(false); } } fast;

#if defined(PCM) || defined(LOCAL)
  #include "lib/dump.hpp"
#else
  #define dump(...) 42
  #define dump_1d(...) 42
  #define dump_2d(...) 42
#endif
//}}}

int dp[2001][2001];

int solve(){
  int n,m;cin>>n>>m;
  vector<int> s(n+1, -1);
  vector<int> t(m+1, -1);
  rep(i,n) cin>>s[i+1];
  rep(i,m) cin>>t[i+1];
  dump(s,t);

  rep(j, m+1) dp[0][j]=1;

  int res=0;
  rep(i, 1, n+1){
    rep(j, m+1){
      dp[i][j] += dp[i][j-1]%MOD;
      if (s[i]==t[j]){
        dp[i][j] += dp[i-1][j-1]%MOD;
      }
      dp[i][j] %= MOD;
    }
    rep(j, m+1){
      dp[i][j] += dp[i-1][j]%MOD;
      dp[i][j] %= MOD;
    }
  }
  // dump_2d(dp, n+1, m+1);

  cout << dp[n][m]%MOD << endl;

  return 0;
}

signed main() { //{{{
#ifdef INPUT_FROM_FILE
  std::ifstream in(infile);
  std::cin.rdbuf(in.rdbuf());
#endif
  solve();
  return 0;
} //}}}