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<stdio.h>
#include<iostream>
#include<algorithm>
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i++)
using namespace::std;

int main(){
	long long int n,t[20000]={0},a[20000]={0},d[20000]={0},f[20000]={0},mx=0,ans=1;
	scanf("%lld",&n);
	REP(i,n)scanf("%lld",&t[i]);
	REP(i,n)scanf("%lld",&a[i]);
	
	REP(i,n)d[i]=min(t[i],a[i]);
	REP(i,n)mx=max(mx,a[i]);
	
	
//	printf("%lld\n",mx);
	long long int i=0,m,k;
	while(i<n){
		m=d[i];
		k=0;
		if(d[i]==mx)k--;
		while(i<n){
			if(d[i]!=d[i+1])break;
			k++;
			i++;
			
		}
		if(k>0){
			REP(c,k){ans=ans*m;
			ans=ans%1000000007;
//			printf("[%lld^%lld]",m,c);
//			printf("(%lld)%lld:",i,ans);
			}
		}
//		printf("\n(%lld)%lld:",i,ans);
		i++;
	}
	printf("%lld\n",ans);
	return 0;
}