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>

int main(){

    int N,i;
    scanf("%d",&N);
    int T[N];
    for(i=0;i<N;i++){
        scanf("%d",&T[i]);
    }
    int A[N];
    for(i=0;i<N;i++){
        scanf("%d",&A[i]);
    }
    int P[N];
    int mid = N/2;
    if(N%2==1){N++;}

    long long res;
    for(i=0;i<N;i++){
        if(i==0 ||(T[i]!=T[i-1])){
            if(A[i]<T[i])
                res=0;
        }
        if(i==N-1 ||(A[i]!=A[i+1])){
            if(A[i]>A[i])
                res=0;
        }
    }
    if(res==0){printf("0"); return 0;}
    for(i=0;i<N;i++){
        if(i==0 ||(T[i]!=T[i-1])){
            P[i]=1;
        }else{
            P[i]=T[i];
        }
    }
    for(i=N-1;i>=0;i--){
        if(i==N-1 ||(A[i+1]!=A[i])){
            P[i]=1;
        }else{
            if(A[i]<P[i]){P[i]=A[i];}
        }
    }
    res=1;
    for(i=0;i<N;i++){
        printf("%d-",P[i]);
        res*=P[i];
    }
    while(res > 1000000007){
        res-=1000000007;
    }
    printf("%d",res);
    return 0;
}