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: Accepted)

#include <bits/stdc++.h>
#include <math.h>
#define _GLIBCXX_DEBUG
#define _LIBCPP_DEBUG 0

using namespace std;
#define ll long long
#define rep(i,n) for (int i = 0; i < n; i++)
#define rrep(i,n) for (int i = n-1; i >= 0; i--)
#define MOD (1000000007)
#define vi vector<int>
#define vl vector<ll>
#define vb vector<bool>
#define vd vector<double>
#define vvi vector<vi>
#define vvl vector<vl>
#define vvb vector<vb>
#define vvd vector<vd>
#define pii pair<int, int>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define pb push_back
#define sz(s) (int)(s.size())

#define mp make_pair
#define all(a) (a).begin(),(a).end()

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

ll gcd(ll a, ll b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

ll keta(ll n){
  string s = to_string(n);
  ll num = s.size();
  return num;
}

const ll INF = 1LL << 60;

const int dh[4] = {1,0,-1,0};
const int dw[4] = {0,1,0,-1};

struct Edge{
  int to;
  int weight;
  Edge(int t, int w) : to(t), weight(w){}
};
using Graph = vector<vector<Edge>>;
using P = pair<ll, int>;

class UnionFind{
public:
  vi Parent, diff_weight;
  UnionFind(int n){
    Parent = vi(n,-1);
    diff_weight = vi(n,0);
  }

  int root(int a){
    if(Parent[a] < 0) return a;
    int r = root(Parent[a]);
    diff_weight[a] += diff_weight[Parent[a]];
    return Parent[a] = r;
  }

  int weight(int a){
    root(a);
    return diff_weight[a];
  }

  bool issame(int a, int b){
    return root(a) == root(b);
  }

  int size(int a){
    return -Parent[root(a)];
  }

  // weight(b) - weight(a) = w となるように繋ぐ
  // 数直線上で a........b となっていてその距離が w 、のようなイメージ
  bool merge(int a, int b, int w = 0){
    w += weight(a); w -= weight(b);

    a = root(a);
    b = root(b);

    if(a == b) return false;

    if(size(a) < size(b)) swap(a,b), w = -w;
    Parent[a] += Parent[b];
    Parent[b] = a;
    diff_weight[b] = w;
    return true;
  }

  int diff(int a, int b){
    return weight(b) - weight(a);
  }
};

vi MP(string s){
  vi A(s.size()+1);
  A[0] = -1;
  int j = -1;
  rep(i,s.size()) {
    while(j >= 0 && s[i] != s[j]) j = A[j];
    j++;
    A[i+1] = j;
  }
  return A;
}

vi Manacher(string s){
  vi R(s.size());
  int i = 0, j = 0;
  while(i < s.size()){
    while(i-j >= 0 && i+j < s.size() && s[i-j] == s[i+j]) ++j;
    R[i] = j;
    int k = 1;
    while(i-k >= 0 && i+k < s.size() && k+R[i-k] < j) R[i+k] = R[i-k], k++;
    i += k; j -= k;
  }
  return R;
}

vi Z_algorithm(string &s){
  vi Z(s.size());
  Z[0] = s.size();
  int i = 1, j = 0;
  while(i < s.size()){
    while(i+j < s.size() && s[j] == s[i+j]) j++;
    Z[i] = j;
    if(j == 0){++i; continue;}
    int k = 1;
    while(i+k < s.size() && k+Z[k] < j) Z[i+k] = Z[k], ++k;
    i += k; j -= k;
  }
  return Z;
}

const int MAX = 1e6+1;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(ll n, ll k){
  if(n >= MAX){
    ll tmp = 1;
    rep(i,k){
      tmp *= (n-i);
      tmp %= MOD;
    }
    return tmp*finv[k]%MOD;
  }

  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll POW(ll x, ll n){
  ll ret = 1;
  if(n < 0){// n < 0 にも対応
    n *= -1;
    x = inv[x];
  }
  while(0 < n){
    if(n%2 == 0){
      x = x*x%MOD;
      n /= 2;
    }
    else{
      ret = ret*x%MOD;
      n--;
    }
  }
  return ret;
}

int dp[2][2020][2020];

int main(){
  int n, m; cin >> n >> m;
  vi s(n+1), t(m+1);
  rep(i,n) cin >> s[i];
  rep(i,m) cin >> t[i];
  // dp[0][i][j]: iを進めるモード
  dp[0][0][0] = 1;
  rep(i,n+1) rep(j,m+1) {
    // dp[0][i][j] (iを進めるモードの状態数) は
    // dp[0][i+1][j] と dp[1][i][j] に配る
    (dp[0][i+1][j] += dp[0][i][j]) %= MOD;
    (dp[1][i][j] += dp[0][i][j]) %= MOD;
    // dp[1][i][j] (j を進めるモード、つまりiを進め終わった状態数) は
    // dp[1][i][j] と dp[0][i+1][j+1] に配る
    (dp[1][i][j+1] += dp[1][i][j]) %= MOD;
    if(s[i] == t[j]) (dp[0][i+1][j+1] += dp[1][i][j]) %= MOD;
  }

  // rep(i,n+1) rep(j,m+1) {
  //   if(j < m+1-1) printf("(%d %d) ", dp[0][i][j], dp[1][i][j]);
  //   else printf("(%d %d)\n", dp[0][i][j], dp[1][i][j]);
  // }

  cout << dp[1][n][m] << endl;
}