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

typedef pair<int,int> pii;
typedef long long ll;
typedef vector<ll> vi;

#define rep(i,j) for(int i=0;i<(j);i++)
#define reps(i,j,k) for(int i=j;i<k;i++)
#define in(i,j,k) ((i)>=(j)&&(i)<=(k))
#define INF (1<<28)
#define pb push_back
#define fs first
#define sc second

const char dx[]={0,1,0,-1},dy[]={1,0,-1,0};
template<class T>
ostream &operator<<(ostream &out, const vector<T> &v){
	out << "{";
	rep(i,v.size()) out << v[i] << ", " ;
	return out << "}" << endl;
}
template<class S,class T>
ostream &operator<<(ostream &out, const pair<S,T> p){
	return out << "(" << p.fs << ", " << p.sc << ")";
}
#define MOD 1000000007
int main(){
	int n;
	cin >> n;
	vi t(n), a(n);
	ll ans = 1;
	rep(i,n){
		cin >> t[i];
		if(i&&t[i]<t[i-1])ans=0;
	}
	rep(i,n){
		cin >> a[i];
		if(i&&a[i]>a[i-1])ans=0;
	}
	vi det(n);
	det[0]=t[0];
	det[n-1]=a[n-1];
	int tmp = 0;
	rep(i,n){
		if(i && t[i]!=t[i-1]){
			if(i==n-1&&t[i]!=a[i]) ans = 0;
			det[i]=t[i];
			if(tmp >= det[i])ans=0;
			tmp = det[i];
		}
		if(det[i]&&det[i]>a[i]) ans = 0;
	}
	tmp=0;
	rep(j,n){
		int i=n-1-j;
		if(i!=n-1 && a[i]!=a[i+1]){
			if(det[i]!=0 && a[i]!=det[i])ans=0;
			det[i] = a[i];
			if(tmp >= det[i]) ans=0;
			tmp = det[i];
		}
	}
	int m;
	rep(i,n){
		int l = max(t[i],a[i]);
		if(i&&m!=l)ans=0;
		m=l;
	}
	rep(i,n){
		if(!det[i]){
			ans *= min(t[i],a[i]);
			ans = ans % MOD;
		}
	}
	cout << ans << endl;
	
	return 0;
}