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 <bits/stdc++.h>

#ifdef _DEBUG
#define debug(x) cerr << "line: " << __LINE__ << ", func: " << __func__ << " ->  " << #x << " = " << x << endl
#else
#define debug(x)
#endif

#define all(s) begin(s), end(s)
#define rall(s) rbegin(s), rend(s)
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define rrep(i, a, b) for (int i = ((a)-1); i >= (b); i--)
#define pb push_back
#define sz(a) int((a).size())
#define put(a) ((cout) << (a) << (endl))
#define fi first
#define se second

using namespace std;

using ll = long long;
using ull = unsigned long long;
using i_i = pair<int, int>;
using ll_ll = pair<ll, ll>;
using d_ll = pair<double, ll>;
using ll_d = pair<ll, double>;
using d_d = pair<double, double>;
template <class T> using vec = vector<T>;

static constexpr ll LL_INF = 1LL << 60;
static constexpr int I_INF = 1 << 28;
static constexpr double PI = static_cast<double>(3.14159265358979323846264338327950288);
static constexpr double EPS = numeric_limits<double>::epsilon();
static constexpr ll MOD = 1000000007;

static map<type_index, const char* const> scanType = {
    {typeid(int), "%d"}, {typeid(ll), "%lld"}, {typeid(double), "%lf"}, {typeid(char), "%c"}};

template <class T> static void scan(vector<T>& v);
[[maybe_unused]] static void scan(vector<string>& v, bool isWord = true);
template <class T> static inline bool chmax(T& a, T b);
template <class T> static inline bool chmin(T& a, T b);
template <class T> static inline T gcd(T a, T b);
template <class T> static inline T lcm(T a, T b);
template <class A, size_t N, class T> static void Fill(A (&arr)[N], const T& val);

ll dp[2005][2005];
ll sum[2005][2005];

int main(int argc, char* argv[]) {
  ll n, m;
  cin >> n >> m;
  vec<ll> s(n), t(m);
  scan(s), scan(t);

  rep(i, 0, n) rep(j, 0, m) {
    if (s[i] == t[j]) {
      dp[i][j] = (i == 0 || j == 0) ? 1 : sum[i - 1][j - 1] + 1;
      dp[i][j] %= MOD;
    }

    sum[i][j] = dp[i][j];
    if (i > 0) sum[i][j] += sum[i - 1][j];
    if (j > 0) sum[i][j] += sum[i][j - 1];
    if (i > 0 && j > 0) sum[i][j] -= sum[i - 1][j - 1];
    sum[i][j] %= MOD;
  }

  cout << sum[n - 1][m - 1] + 1 << endl;
  return 0;
}

template <class T> static void scan(vector<T>& v) {
  auto tFormat = scanType[typeid(T)];
  for (T& n : v) {
    scanf(tFormat, &n);
  }
}

static void scan(vector<string>& v, bool isWord) {
  if (isWord) {
    for (auto& n : v) {
      cin >> n;
    }
    return;
  }

  int i = 0, size = v.size();
  string s;
  getline(cin, s);

  if (s.size() != 0) {
    i++;
    v[0] = s;
  }

  for (; i < size; ++i) {
    getline(cin, v[i]);
  }
}

template <class T> inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T> inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T> inline T gcd(T a, T b) { return __gcd(a, b); }

template <class T> inline T lcm(T a, T b) { return (a * b) / gcd(a, b); }

template <class A, size_t N, class T> void Fill(A (&arr)[N], const T& val) {
  std::fill((T*)arr, (T*)(arr + N), val);
}