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(void){
    int i,n;
    scanf("%d",&n);
    int t[n+2],a[n+2],h[n+2],max=0,maxi=0;
    t[0]=0;
    t[n+1]=0;
    a[0]=0;
    a[n+1]=0;
    for(i=1;i<=n;i++){
        scanf("%d",&t[i]);
        h[i]=0;
    }
    for(i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }

    for(i=0;i<n;i++){
        if(t[i]<t[i+1]){
            h[i+1]=t[i+1];
            if(max<h[i+1]){
                max=h[i+1];
                maxi=i+1;
            }
        }
    }
    for(i=n+1;maxi<i-1;i--){
        if(a[i]<a[i-1]){
            h[i-1]=a[i-1];
        }
    }
    if(t[maxi]!=a[maxi]){
        printf("0\n");
        return 0;
    }
    if(n==1){
        printf("1\n");
        return 0;
    }
    int j=0,k,z;
    unsigned long long x=1,y=0;
    for(i=1;i<=maxi;i++){
        if(h[i]==0){
            j++;
        }
        else{
            if(j!=0){
            for(k=0;k<j;k++){
                x*=z;
            }
            y+=x;
            x=1;
            j=0;
            }
            z=h[i];
        }
    }
    j=0;
    for(i=n;maxi<=i;i--){
        if(h[i]==0){
            j++;
        }
        else{
            if(j!=0){
            for(k=0;k<j;k++){
                x*=z;
            }
            y+=x;
            x=1;
            j=0;
            }
            z=h[i];
        }
    }
    y=y%1000000007;
    printf("%llu\n",y);
    return 0;
}