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: Accepted)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define pb push_back
#define INF (1e9+1)
//#define INF (1LL<<59)


ll solve(pii a,pii b){
	if(a.first<=b.first&&b.second<=a.second){
		return b.second-b.first+1;
	}
	else if(b.first<=a.first&&a.second<=b.second){
		return a.second-a.first+1;
	}
	else if(a.first<=b.first&&b.first<=a.second){
		return a.second-b.first+1;
	}
	else if(b.first<=a.first&&a.first<=b.second){
		return b.second-a.first+1;
	}
	else return 0;
}

const ll MOD = (1e9)+7;
int main(){
	int n;
	cin>>n;
	vector<int> t(n);
	vector<int> a(n);
	rep(i,n)cin>>t[i];
	rep(i,n)cin>>a[i];
	
	vector<pii> tt(n),aa(n);
	
	tt[0] = pii(t[0],t[0]);
	for(int i=1;i<n;i++){
		if(t[i]>t[i-1]){
			tt[i] = pii(t[i],t[i]);
		}
		else{
			tt[i] = pii(1,t[i]);
		}
	}
	
	
	aa[n-1] = pii(a[n-1],a[n-1]);
	for(int i=n-2;i>=0;i--){
		if(a[i]>a[i+1]){
			aa[i] = pii(a[i],a[i]);
		}
		else{
			aa[i] = pii(1,a[i]);
		}
	}
	
	
	ll ans = 1;
	rep(i,n){
		if(tt[i].second<aa[i].first || aa[i].second<tt[i].first){
			cout<<0<<endl;
			return 0;
		}
		pii x=tt[i],y=aa[i];
		ll  tmp = solve(x,y)%MOD;
		ans = (ans*tmp)%MOD;
	}
	cout<<ans<<endl;
}