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<string.h>
#include<math.h>

#define scanl(x) scanf("%ld", &x)
#define scans(x) scanf("%s", x)

#define INF 1000000000
#define M 1000000007

#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

int main() {
	long N;
	long T[100001];
	long A[100001];
	int i;
	long m;
	long x[100001];
	long ans = 1;
	
	scanl(N);
	for(i=0;i<N;i++) {
		scanl(T[i]);
	}
	for(i=0;i<N;i++) {
		scanl(A[i]);
	}
	
	m = T[0];
	x[0] = T[0];
	for(i=1;i<N;i++) {
		if(m != T[i]) {
			x[i] = T[i];
			m = T[i];
			if(T[i] > A[i]) ans = 0;
		} else {
			x[i] = INF;
		}
	}
	m = A[N-1];
	x[N-1] = A[N-1];
	for(i=N-2;i>=0;i--) {
		if(m != A[i]) {
			x[i] = min(x[i], A[i]);
			m = A[i];
			if(A[i] > T[i]) ans = 0;
		}
	}
	
	for(i=0;i<N;i++) {
		if(x[i] == INF) {
			ans *= min(T[i], A[i]);
			ans %= M;
		}
	}
	printf("%ld\n", ans);
	
	return 0;
}