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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cassert>
#define REP(i,s,n) for(int i=s;i<n;++i)
#define rep(i,n) REP(i,0,n)
using namespace std;
long long ans;
long long n,t[100000],a[100000];
long long subans[100000];
bool fix[100000];
int main() {
  cin >> n;
  rep(i,n) cin >> t[i];
  rep(i,n) cin >> a[i];
  subans[0] = 1;
  REP(i,1,n){
    if(t[i]>t[i-1]) subans[i] = 1;
    else subans[i] = t[i];
  }
  if(t[n-1]<a[n-1]){
    cout << 0 << endl;
    return 0;
  }
  subans[n-1] = 1;
  for(int i = n-2;i>=0;--i){
    if(t[n-1]<a[i]){
      cout << 0 << endl;
      return 0;
    }
    if(a[i]>a[i+1]) subans[i] = 1;
    else if(subans[i]>a[i]) subans[i] = a[i];
  }
  ans = 1;
  rep(i,n){
    ans*=subans[i];
    ans%=1000000007;
    //    cout << ans << endl;
  }
  cout << ans << endl;
  return 0;
}