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: Memory Limit Exceeded)

#if 1
#include <iostream>
#include <cmath>
#include <string>
#include <vector>
#include <climits>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>
#include <functional>
#include <set>
#include <map>
#include <numeric>
#include <cassert>
#include <iomanip>
#define _SCL_SECURE_NO_WARNINGS
#include "boost/multiprecision/cpp_int.hpp"

#pragma warning (disable:4244)


using namespace std;

//#define int boost::multiprecision::int128_t
#define int long long
constexpr long long MOD = 1000000007LL;
//constexpr int MOD = 998244353;
constexpr long long INF = 1145141919810893LL;

//
#if 1
struct Edge {
	int next;
	int w = 1;
};
template<class T>
struct Vertex {
	std::vector<Edge>edges;
	T val = {};
};
template<class T>
struct Graph {
	std::vector<Vertex<T>> vertex;
public:
	Graph(size_t n = 0) :vertex(n) {
	}
	void setArray(int u, int v, int w = 1) {
		vertex[u].edges.push_back(Edge{ v ,w });
	}
	void setConnect(int u, int v, int w = 1) {
		setArray(u, v, w);
		setArray(v, u, w);
	}
	T& val(int u) {
		return vertex[u].val;
	}
	void dfsImpl(int pos, int prev, std::function<void(int now, int next, int w)> func) {
		for (auto& e : vertex[pos].edges) {
			if (e.next == prev) continue;
			func(pos, e.next, e.w);
			dfsImpl(e.next, pos, func);
		}
	}
	void dfs(int pos, std::function<void(int now, int next, int w)> func) {
		dfsImpl(pos, -1, func);
	}
	//graph.dfs(0, [](Vertex<int>& now, Vertex<int>& next, int w) {});
	pair<int, int> radiusImpl(int pos) {
		int farestID = -1;
		int far = -1;
		val(pos) = 0;
		dfs(pos, [&](int now, int next, int w) {
			val(next) = 1 + val(now);
			if (val(next) > far) {
				far = val(next);
				farestID = next;
			}
			});
		return { farestID, far };
	}
	int radius() {
		if (vertex.size() <= 1)return 0;
		auto res = radiusImpl(0);
		res = radiusImpl(res.first);
		return res.second;
	}
};
#endif

//////////////////////////////////
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }

std::vector<int> divisor(int n) {
	std::vector<int> ret;
	for (int i = 1; i * i <= n; ++i) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) {
				ret.push_back(n / i);
			}
		}
	}
	return ret;
}

bool isPrime(int n) {
	if (n <= 1)return false;
	if (n == 2)return true;
	if (n % 2 == 0)return false;
	for (int i = 3; i * i <= n; i += 2) {
		if (n % i == 0)return false;
	}
	return true;
}

////////////////////////
#if 1
//二項係数、int128は使わなく64で十分
class Combination {
	std::vector<long long>fac, finv, inv;
public:
	Combination(long long N) :fac(N + 1), finv(N + 1), inv(N + 1) {
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[1] = 1;
		for (long long i = 2; i < N + 1; 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;
		}
	}
	long long get(long long n, long long k) {
		if (n < k) return 0;
		if (n < 0 || k < 0) return 0;
		return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
	}
};

#endif
#if 1

int modInv(int a) {
	int b = MOD, u = 1, v = 0;
	while (b) {
		int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= MOD;
	if (u < 0) u += MOD;
	return u;
}


void modify(int& n, int mod = MOD) {
	if (n < 0) {
		n %= mod;
		n += mod;
	}
	n %= mod;
}

#endif

//文字列の置き換え(遅い？)
int replace(std::string* str, const std::string& old_, const std::string& new_) {
	std::string& String1 = *str;
	//String1.reserve(str->size() * new_.size() / old_.size() + 1);
	std::string::size_type  Pos(String1.find(old_));
	int result = 0;
	while (Pos != std::string::npos) {
		result++;
		String1.replace(Pos, old_.length(), new_);
		Pos = String1.find(old_, Pos + new_.length());
	}
	return result;
}

//MODで割ったあまりの演算
struct Rational {
	int r;
	Rational(int rr) :r(rr) {
	}
	operator int() {
		return r;
	}
	Rational operator+(const Rational& other)const {
		Rational res(0);
		res.r = r + other.r;
		modify(res.r);
		return res;
	}
	Rational operator-(const Rational& other)const {
		Rational res(0);
		res.r = r - other.r;
		modify(res.r);
		return res;
	}
	Rational operator*(const Rational& other) const {
		Rational res(0);
		res.r = r * other.r;
		modify(res.r);
		return res;
	}
	Rational operator/(const Rational& other)const {
		Rational res(0), res1(0);
		res = *this;
		res1.r = modInv(other.r);
		res = (*this) * res1;
		modify(res.r);
		return res;
	}
	void operator+=(const Rational& other) {
		*this = *this + other;
	}
	void operator-=(const Rational& other) {
		*this = *this - other;
	}
	void operator*=(const Rational& other) {
		*this = *this * other;
	}
	void operator/=(const Rational& other) {
		*this = *this / other;
	}
};

Rational pow(Rational r, int N) {
	if (N == 0)return Rational(1);
	if (N % 2 == 1)return r * pow(r, N - 1);
	Rational tmp = pow(r, N / 2);
	return tmp * tmp;
}
Rational operator"" _r(unsigned long long val) {
	return Rational(val);
}

/////////////////////

template <class T>
void SORT(T& t) {
	std::sort(t.begin(), t.end());
}

template <class T, class U>
void FILL(T& t, const U& val) {
	std::fill(t.begin(), t.end(), val);
}

constexpr double pi = 3.141592653589793238462;
//////////////////

#define LOADVEC(type,name,N) std::vector<type>name(N); \
for (int nnn = 0; nnn < N; ++nnn) { \
	cin >> name[nnn]; \
}

#define LOADVEC2(type,name0,name1,N) std::vector<type>name0(N),name1(N); \
for (int nnn = 0; nnn < N; ++nnn) { \
	cin >> name0[nnn];cin >> name1[nnn]; \
}

#define LOADVEC3(type,name0,name1,name2,N) std::vector<type>name0(N),name1(N),name2(N); \
for (int nnn = 0; nnn < N; ++nnn) { \
	cin >> name0[nnn];cin >> name1[nnn];cin >> name2[nnn]; \
}

#define LOAD(type,name) type name; \
cin >> name;

void proc();

signed main() {
	ios::sync_with_stdio(false);
	cout << std::setprecision(15);
	proc();
	return 0;
}

/*
--------------------------------------------------------
--------------------------------------------------------
---------------    template      ----------------------
--------------------------------------------------------
--------------------------------------------------------
*/



void proc() {
	LOAD(int, N);
	LOAD(int, M);
	LOADVEC(int, S, N);
	LOADVEC(int, T, M);
	vector<int>dp0(N + 1);
	vector<int>dp1(N + 1);
	vector<vector<int>>dp2(N + 1);
	for (int i = 0; i < N + 1; ++i) {
		dp2[i].resize(114514);
		FILL(dp2[i], 0);
		dp2[i][T[0]] = 1;
	}
	FILL(dp1, 1);
	for (int i = 1; i < M + 1; ++i) {
		dp0 = dp1;
		for (int j = 0; j < N + 1; ++j) {
			if (j == 0) {
				dp1[j] = 1;
				continue;
			}
			dp1[j] = dp1[j - 1];
			//if (S[j - 1] == T[i - 1]) {
			dp1[j] += dp2[j - 1][S[j - 1]];
			//}
			modify(dp1[j]);
			//
		}
		if (i != M) {
			for (int j = 0; j < N + 1; ++j) {
				dp2[j][T[i]] += dp1[j];
				modify(dp2[j][T[i]]);
			}
		}
	}
	cout << dp1[N] << endl;
}


#endif