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 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#include <math.h>

#define MOD 1000000007
#define REP(i,n) for ((i)=0;i<(n);i++)
#define SET(a,c) memset(a,c,sizeof a)
#define CLR(a) memset(a,0,sizeof a)
#define MIN(a,b) (a>b?b:a)
#define MAX(a,b) (a>b?a:b)
#define LL long long
#define ULL unsigned long long

int compare_ll(const void *a, const void *b){
	return *(ULL*)b - *(ULL*)a;
}

int compare_int(const void *a, const void *b)
{
    return *(int*)b - *(int*)a;
}

int main(void) {

	ULL N,T[100001],A[100001],ans,i,j;
	int flag = 0;
	ans = 1;
	scanf("%lld", &N);
	for(i = 0; i<N;i++){
		scanf("%lld", &T[i]);
	}

	for(i = 0; i<N;i++){
		scanf("%lld", &A[i]);
	}
	T[i] = 0;
	A[i] = 0;

	for(i = 1;i<N;i++){
		if(T[i]!=T[i-1]){
			if(T[i]>A[i]){
				flag = 1;
				break;
			}
		}
	}

	for(i = N-1;i==0;i--){
		if(A[i-1]!=A[i]){
			if(A[i-1]>T[i-1]){
				flag = 1;
				break;
			}
		}
	}

	for(i = 1;i<N;i++){
		if(T[i]==T[i-1]){
			if(A[i]==A[i+1]){
				if(A[i] > T[i]){
					ans *= T[i];
				}else if(T[i] == A[i]){
					ans *= A[i];
				}else{
					ans *= A[i];
				}
				ans %= MOD;
			}
		}
	}

	if(flag == 1){
		ans = 0;
	}

	printf("%lld", ans);


	return 0;
}

/* 最後の改行忘れるな！！ */