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>

using namespace std;

using ll = long long;
using Vi = vector<int>;
using Vl = vector<ll>;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;

constexpr int I_INF = (1 << 30);
constexpr ll L_INF = (1ll << 62);

//==================================

template <std::int_fast64_t Modulus>
class ModInt {
    using i64 = std::int_fast64_t;

    // friend
    template <std::int_fast64_t _MOD>
    friend std::ostream& operator<<(std::ostream& os, const ModInt<_MOD>& rhs);

private:
    i64 a;

public:
    /**
     * コンストラクタ
     */
    constexpr ModInt(const i64 x = 0) noexcept : a(x % Modulus) {}

    constexpr i64& value() noexcept { return a; }
    constexpr const i64& value() const noexcept { return a; }

    constexpr ModInt operator+(const ModInt& rhs) const noexcept {
        return ModInt(*this) += rhs;
    }
    constexpr ModInt operator-(const ModInt& rhs) const noexcept {
        return ModInt(*this) -= rhs;
    }
    constexpr ModInt operator*(const ModInt& rhs) const noexcept {
        return ModInt(*this) *= rhs;
    }
    constexpr ModInt operator/(const ModInt& rhs) const noexcept {
        return ModInt(*this) /= rhs;
    }

    constexpr ModInt& operator+=(const ModInt rhs) noexcept {
        a += rhs.a;
        if (a >= Modulus) {
            a -= Modulus;
        }
        return *this;
    }
    constexpr ModInt& operator-=(const ModInt rhs) noexcept {
        if (a < rhs.a) {
            a += Modulus;
        }
        a -= rhs.a;
        return *this;
    }
    constexpr ModInt& operator*=(const ModInt rhs) noexcept {
        a = a * rhs.a % Modulus;
        return *this;
    }
    constexpr ModInt& operator/=(ModInt rhs) noexcept {
        i64 exp = Modulus - 2;
        while (exp) {
            if (exp & 1) {
                *this *= rhs;
            }
            rhs *= rhs;
            exp >>= 1;
        }
        return *this;
    }

    constexpr ModInt& operator++() noexcept {
        *this += 1;
        return *this;
    }
    constexpr ModInt operator++(int) noexcept {
        ModInt res = *this;
        *this += 1;
        return res;
    }
    constexpr ModInt& operator--() noexcept {
        *this -= 1;
        return *this;
    }
    constexpr ModInt operator--(int) noexcept {
        ModInt res = *this;
        *this -= 1;
        return res;
    }

    constexpr bool operator<(const ModInt& rhs) const noexcept {
        return a < rhs.a;
    }
    constexpr bool operator>(const ModInt& rhs) const noexcept {
        return a > rhs.a;
    }
    constexpr bool operator<=(const ModInt& rhs) const noexcept {
        return a <= rhs.a;
    }
    constexpr bool operator>=(const ModInt& rhs) const noexcept {
        return a >= rhs.a;
    }
    constexpr bool operator==(const ModInt& rhs) const noexcept {
        return a == rhs.a;
    }

    constexpr ModInt pow(i64 k) const noexcept {
        ModInt res = 1;
        ModInt tmp = *this;
        for (; k > 0; k >>= 1) {
            if (k & 1) res *= tmp;
            tmp *= tmp;
        }
        return res;
    }

    constexpr ModInt inverse() const noexcept {
        i64 q = a;
        i64 b = Modulus, u = 1, v = 0;
        while (b) {
            i64 t = q / b;
            q -= t * b;
            swap(q, b);
            u -= t * v;
            swap(u, v);
        }
        u %= Modulus;
        if (u < 0) u += Modulus;
        return u;
    }
};

// ModIntでcoutするための関数
template <std::int_fast64_t Modulus>
std::ostream& operator<<(std::ostream& os, const ModInt<Modulus>& rhs) {
    os << rhs.a;
    return os;
}

constexpr ll MOD = (ll)(1e9 + 7);

ModInt<MOD> dp[2010][2010];
ModInt<MOD> sum[2010][2010];
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M;
    cin >> N >> M;
    Vi S(N + 1), T(M + 1);
    for (int i = 0; i < N; ++i) {
        cin >> S[i];
    }
    for (int i = 0; i < M; ++i) {
        cin >> T[i];
    }

    for (int i = 1; i <= N + 1; ++i) {
        for (int j = 1; j <= M + 1; ++j) {
            sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + dp[i - 1][j - 1];
            if (S[i - 1] == T[j - 1]) {
                dp[i][j] = sum[i][j] + 1;
            }
        }
    }

    cout << sum[N + 1][M + 1] + 1 << endl;

    return 0;
}