Problem:
Score : 400 points

Problem Statement

Mountaineers Mr. Takahashi and Mr. Aoki recently trekked across a certain famous mountain range.
The mountain range consists of N mountains, extending from west to east in a straight line as Mt. 1, Mt. 2, ..., Mt. N.
Mr. Takahashi traversed the range from the west and Mr. Aoki from the east.

The height of Mt. i is h_i, but they have forgotten the value of each h_i.
Instead, for each i (1 ≤ i ≤ N), they recorded the maximum height of the mountains climbed up to the time they reached the peak of Mt. i (including Mt. i).
Mr. Takahashi's record is T_i and Mr. Aoki's record is A_i.

We know that the height of each mountain h_i is a positive integer.
Compute the number of the possible sequences of the mountains' heights, modulo 10^9 + 7.

Note that the records may be incorrect and thus there may be no possible sequence of the mountains' heights.
In such a case, output 0.

Constraints

1 ≤ N ≤ 10^5

1 ≤ T_i ≤ 10^9

1 ≤ A_i ≤ 10^9

T_i ≤ T_{i+1} (1 ≤ i ≤ N - 1)

A_i ≥ A_{i+1} (1 ≤ i ≤ N - 1)

Input

The input is given from Standard Input in the following format:

N
T_1 T_2 ... T_N
A_1 A_2 ... A_N

Output

Print the number of possible sequences of the mountains' heights, modulo 10^9 + 7.

Sample Input 1

5
1 3 3 3 3
3 3 2 2 2

Sample Output 1

4

The possible sequences of the mountains' heights are:

1, 3, 2, 2, 2

1, 3, 2, 1, 2

1, 3, 1, 2, 2

1, 3, 1, 1, 2

for a total of four sequences.

Sample Input 2

5
1 1 1 2 2
3 2 1 1 1

Sample Output 2

0

The records are contradictory, since Mr. Takahashi recorded 2 as the highest peak after climbing all the mountains but Mr. Aoki recorded 3.

Sample Input 3

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5

Sample Output 3

884111967

Don't forget to compute the number modulo 10^9 + 7.

Sample Input 4

1
17
17

Sample Output 4

1

Some mountain ranges consist of only one mountain.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//include
//------------------------------------------
#include <bits/stdc++.h>
using namespace std;

//typedef
//------------------------------------------
using LL = int64_t;
using VL = vector<LL>;
using VVL = vector<VL>;
using VS = vector<string>;
using PLL = pair<LL, LL>;

//container util
//------------------------------------------
#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)
#define rwhole(f,x,...) ([&](decltype((x)) whole) { return (f)(rbegin(whole), rend(whole), ## __VA_ARGS__); })(x)
#define EACH(i,c) for(decltype((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) whole(sort, c)
#define RSORT(c) rwhole(sort, c)

//constant
//--------------------------------------------
constexpr double EPS = 1e-10;
constexpr double PI  = M_PI;
constexpr int MOD = 1000000007;

// grid
//--------------------------------------------
VL dx = {0, 1, 0, -1};
VL dy = {1, 0, -1, 0};
VL dx2 = {-1, 0, 1, -1, 1, -1, 0, 1};
VL dy2 = {-1, -1, -1, 0, 0, 1, 1, 1};

//debug
//--------------------------------------------
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

//IO accelerate
//--------------------------------------------
struct InitIO {
    InitIO() {
        cin.tie(nullptr);
        ios_base::sync_with_stdio(false);
        cout << fixed << setprecision(30);
    }
} init_io;

//template
//--------------------------------------------
// declaretion
template<typename T> istream& operator >>(istream& is, vector<T>& vec);
template<typename T1, typename T2> ostream& operator <<(ostream& os, const pair<T1, T2>& p);
template<typename T> ostream& operator <<(ostream& os, const vector<T>& vec);
template<typename T> ostream& operator <<(ostream& os, const vector<vector<T>>& vv);
template<typename T> vector<T> make_v(size_t a);
template<typename T,typename... Ts> auto make_v(size_t a,Ts... ts);
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type fill_v(T &t,const V &v);
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type fill_v(T &t,const V &v);

// implementation
template<typename T>
istream& operator >>(istream& is, vector<T>& vec) {
	for(T& x: vec) is >> x;
	return is;
}
template<typename T1, typename T2>
ostream& operator <<(ostream& os, const pair<T1, T2>& p) {
	os << p.first << "," << p.second;
	return os;
}
template<typename T>
ostream& operator <<(ostream& os, const vector<T>& vec) {
	for(int i=0; i<vec.size(); i++){
		os << vec[i] << ( i+1 == vec.size() ? "" : "\t" );
	}
	return os;
}
template<typename T>
ostream& operator <<(ostream& s, const vector<vector<T>>& vv) {
	for (int i = 0; i < vv.size(); ++i) {
		s << vv[i] << endl;
	}
	return s;
}

// 多重vector
// auto dp=make_v<int>(4,h,w) みたいに使える
template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
	return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

// 多重vectorのためのfill
// fill_v(dp,0) みたいに使える
template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
	for(auto &e:t) fill_v(e,v);
}

template<class T,class U> void chmax(T&a,U b){if(a<b)a=b;}
template<class T,class U> void chmin(T&a,U b){if(b<a)a=b;}
template<typename T> T gcd(T a, T b) { return b?gcd(b,a%b):a;}
template<typename T> T lcm(T a, T b) { return a/gcd(a,b)*b;}

//main code

// auto mod int
//
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
// const int MOD = 1000000007;
struct mint {
	LL x; // typedef long long LL;
	constexpr mint(LL x=0):x((x%MOD+MOD)%MOD){}
	constexpr mint operator-() const { return mint(-x);}
	constexpr mint& operator+=(const mint& a) {
		if ((this->x += a.x) >= MOD) this->x -= MOD;
		return *this;
	}
	constexpr mint& operator-=(const mint& a) {
		if ((this->x += MOD-a.x) >= MOD) this->x -= MOD;
		return *this;
	}
	constexpr mint& operator*=(const mint& a) {
		(this->x *= a.x) %= MOD;
		return *this;
	}
	constexpr mint operator+(const mint& a) const {
		mint res(*this);
		return res+=a;
	}
	constexpr mint operator-(const mint& a) const {
		mint res(*this);
		return res-=a;
	}
	constexpr mint operator*(const mint& a) const {
		mint res(*this);
		return res*=a;
	}
	constexpr mint pow(LL t) const {
		if (!t) return 1;
		mint a = pow(t>>1);
		a *= a;
		if (t&1) a *= *this;
		return a;
	}

	// for prime MOD
	constexpr mint inv() const {
		return pow(MOD-2);
	}
	constexpr mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	constexpr mint operator/(const mint& a) const {
		mint res(*this);
		return res/=a;
	}

	constexpr bool operator==(const mint& a) const {
		return this->x == a.x;
	}
	constexpr bool operator!=(const mint& a) const {
		return this->x != a.x;
	}
	constexpr bool operator<(const mint& a) const {
		return this->x < a.x;
	}

	friend istream& operator>>(istream& is, mint& m) {
		return is >> m.x;
	}
	friend ostream& operator<<(ostream& os, const mint& m) {
		return os << m.x;
	}
};
typedef vector<mint> VM;
typedef vector<VM> VVM;

int main(int argc, char *argv[])
{
	LL n;
	cin >> n;
	VL t(n), a(n);
	cin >> t >> a;
	if(*max_element(t.begin(), t.end()) != *max_element(a.begin(), a.end())) {
		cout << 0 << endl;
		return 0;
	}
	VL num(n,1);
	for (int i = 1; i < n-1; i++) {
		if (t[i-1] != t[i] and a[i+1] != a[i] and t[i] != a[i]) {
			cout << 0 << endl;
			return 0;
		}
		if (t[i-1] == t[i] and a[i+1] == a[i]) {
			num[i] = min(t[i], a[i]);
		}
	}
	mint ans = 1;
	for (int i = 0; i < n; i++) {
		ans *= num[i];
	}
	cout << ans << endl;
	return 0;
}