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: Accepted)

#include<stdio.h>
int MIN(int a,int b){return a<b?a:b;}
int main(){
  int n,i,d[2][100010],max,f[100010]={};
  long long s=1,M=1e9+7;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&d[0][i]);
    if(i&&d[0][i-1]<d[0][i])f[i]|=1;
  }
  for(i=0;i<n;i++)scanf("%d",&d[1][i]);
  f[n-1]=2;
  for(i=n-1;i--;){
    if(d[1][i+1]<d[1][i])f[i]|=2;
  }
  for(i=1;i<n-1;i++){
    if(f[i]==0)s=(s*MIN(d[0][i],d[1][i]))%M;
    if(f[i]==1&&d[0][i]>d[1][i])s*=0;
    if(f[i]==2&&d[0][i]<d[1][i])s*=0;
    if(f[i]==3&&d[0][i]-d[1][i])s*=0;
  }
  if(d[0][n-1]-d[1][0])s*=0;
  printf("%lld\n",s);
  return 0;
}