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.

Examples

Input

5
1 3 3 3 3
3 3 2 2 2


Output

4


Input

5
1 1 1 2 2
3 2 1 1 1


Output

0


Input

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5


Output

884111967


Input

1
17
17


Output

1
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)

int main() {
  int mod=1000000007;
  int n;
  cin >> n;
  vector<int>a(n);
  vector<int>b(n);
  rep(i,n){
    cin>>a.at(i);
  }
  rep(i,n){
    cin>>b.at(i);
  }
  bool ch=true;
  rep(i,n-1){
    if(a.at(i+1)!=a.at(i)&&a.at(i+1)>b.at(i+1)){
      ch=false;
    }
    if(b.at(i+1)!=b.at(i)&&a.at(i)<b.at(i)){
      ch=false;
    }
  }
  if(a.at(n-1)!=b.at(0)){
    ch=false;
  }
  if(ch){
    long ans=1;
    rep(i,n-2){
      if(a.at(i)==a.at(i+1)&&b.at(i+2)==b.at(i+1)){
        ans*=min(a.at(i+1),b.at(i+1));
        ans%=mod;
      }
    }
    cout<<ans<<endl;
  }
  else{
    cout<<0<<endl;
  }
  
}