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: Runtime Error)

#include <stdio.h>
int main(void){
    // Here your code !
    
    int n, t[10000],a[1000], ans1=0;
    int ans2=0;
    char T,A;
    int i,j;
    
    scanf("%d",&n);
    gets(T);
    gets(A);
    
    strtok(T," ");
      while ( T != NULL ){
          i=i+1;
      atoi("%s", t[i]);
      T = strtok( NULL, " " );
   }
   
    strtok(A," ");
      while ( A != NULL ){
          i=i+1;
      atoi("%s", a[i]);
      A = strtok( NULL, " " );
   }

    
    if(t[n]!=a[1]){
        printf("0");
        
    }else if(n==1){
        printf("1");
    }else{
    
    for(j=1;j<=n-1;j++){
        if(t[j]==t[j-1]){
            ans1=j*j+ans1;
        }
    }
    
     for(j=1;j<=n-1;j++){
        if(a[j]!=a[j+1]){
            ans2=j*j+ans2;
        }
    }
    
    if(ans1>ans2){
        
        ans2=ans2%(1000000000+7);
        printf("%d",ans2);
    }else{
        
        ans1=ans1%(1000000000+7);
        printf("%d",ans1);
    }
    }
}