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 <stdint.h>
#include <stdlib.h>

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
using namespace std;
using default_counter_t = int64_t;

// macro
#define let auto const &
#define overload4(a, b, c, d, name, ...) name
#define rep1(n) \
  for (default_counter_t i = 0, end_i = default_counter_t(n); i < end_i; ++i)
#define rep2(i, n)                                                           \
  for (default_counter_t i = 0, end_##i = default_counter_t(n); i < end_##i; \
       ++i)
#define rep3(i, a, b)                                    \
  for (default_counter_t i       = default_counter_t(a), \
                         end_##i = default_counter_t(b); \
       i < end_##i; ++i)
#define rep4(i, a, b, c)                                 \
  for (default_counter_t i       = default_counter_t(a), \
                         end_##i = default_counter_t(b); \
       i < end_##i; i += default_counter_t(c))
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)

#define rrep1(n) \
  for (default_counter_t i = default_counter_t(n) - 1; i >= 0; --i)
#define rrep2(i, n) \
  for (default_counter_t i = default_counter_t(n) - 1; i >= 0; --i)
#define rrep3(i, a, b)                                         \
  for (default_counter_t i         = default_counter_t(b) - 1, \
                         begin_##i = default_counter_t(a);     \
       i >= begin_##i; --i)
#define rrep4(i, a, b, c)                                          \
  for (default_counter_t                                           \
           i = (default_counter_t(b) - default_counter_t(a) - 1) / \
                   default_counter_t(c) * default_counter_t(c) +   \
               default_counter_t(a),                               \
           begin_##i = default_counter_t(a);                       \
       i >= begin_##i; i -= default_counter_t(c))
#define rrep(...) \
  overload4(__VA_ARGS__, rrep4, rrep3, rrep2, rrep1)(__VA_ARGS__)

#define ALL(f, c, ...)                                           \
  (([&](decltype((c)) cccc) {                                    \
    return (f)(std::begin(cccc), std::end(cccc), ##__VA_ARGS__); \
  })(c))

// function
template <class C>
constexpr C &Sort(C &a) {
  std::sort(std::begin(a), std::end(a));
  return a;
}
template <class C>
constexpr auto &Min(C const &a) {
  return *std::min_element(std::begin(a), std::end(a));
}
template <class C>
constexpr auto &Max(C const &a) {
  return *std::max_element(std::begin(a), std::end(a));
}
template <class C>
constexpr auto Total(C const &c) {
  return std::accumulate(std::begin(c), std::end(c), C(0));
}
template <typename T>
auto CumSum(std::vector<T> const &v) {
  std::vector<T> a(v.size() + 1, T(0));
  for (std::size_t i = 0; i < a.size() - size_t(1); ++i) a[i + 1] = a[i] + v[i];
  return a;
}
template <typename T>
constexpr bool ChMax(T &a, T const &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
constexpr bool ChMin(T &a, T const &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}

void In(void) { return; }

template <typename First, typename... Rest>
void In(First &first, Rest &... rest) {
  cin >> first;
  In(rest...);
  return;
}

template <class T, typename I>
void VectorIn(vector<T> &v, const I n) {
  v.resize(size_t(n));
  rep(i, v.size()) cin >> v[i];
}

void Out(void) {
  cout << "\n";
  return;
}

template <typename First, typename... Rest>
void Out(First first, Rest... rest) {
  cout << first << " ";
  Out(rest...);
  return;
}

constexpr auto yes(const bool c) { return c ? "yes" : "no"; }
constexpr auto Yes(const bool c) { return c ? "Yes" : "No"; }
constexpr auto YES(const bool c) { return c ? "YES" : "NO"; }

#ifdef USE_STACK_TRACE_LOGGER
#ifdef __clang__
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Weverything"
#include <glog/logging.h>
#pragma clang diagnostic pop
#endif  //__clang__
#endif  // USE_STACK_TRACE_LOGGER

template <int64_t Mod>
struct ModInt {
  int64_t x;

  ModInt() : x(0) {}

  ModInt(int64_t y) : x(y >= 0 ? y % Mod : (Mod - (-y) % Mod) % Mod) {}

  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= Mod) x -= Mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if ((x += Mod - p.x) >= Mod) x -= Mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = int64_t(1LL * x * p.x % Mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.Inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }

  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

  bool operator==(const ModInt &p) const { return x == p.x; }

  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt Inverse() const {
    int64_t a = x, b = Mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt Pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<Mod>(t);
    return (is);
  }

  static int64_t GetMod() { return Mod; }
};

signed main(int argc, char *argv[]) {
  (void)argc;
#ifdef USE_STACK_TRACE_LOGGER
  google::InitGoogleLogging(argv[0]);
  google::InstallFailureSignalHandler();
#else
  (void)argv;
#endif  // USE_STACK_TRACE_LOGGER
  int64_t N, M;

  In(N, M);

  vector<string> S(N + 1), T(M + 1);
  rep(i, N) In(S[i]);
  rep(i, M) In(T[i]);

  using Mi = ModInt<1000000007>;

  vector<vector<Mi>> dp(N + 1, vector<Mi>(M + 1));
  vector<vector<Mi>> cum(N + 1, vector<Mi>(M + 1, 0));

  rep(i, N + 1) {
    dp[i][0]  = 0;
    cum[i][0] = 1;
  }
  rep(j, M + 1) {
    dp[0][j]  = 0;
    cum[0][j] = 1;
  }

  cum[0][0] = 1;
  dp[0][0]  = 1;

  rep(i, 1, N + 1) {
    rep(j, 1, M + 1) {
      if (S[i - 1] != T[j - 1]) {
        dp[i][j] = 0;
      } else {
        dp[i][j] = cum[i - 1][j - 1];
      }
      cum[i][j] = dp[i][j] + cum[i][j - 1] + cum[i - 1][j] - cum[i - 1][j - 1];
    }
  }

  cout << cum[N][M] << endl;

  return EXIT_SUCCESS;
}