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 <bits/stdc++.h>

using namespace std;
typedef pair<int,int> pii;
typedef long long int ll;

#define INF 1 << 30
#define REP(i,n) for(ll i=0; i<(int)(n); i++)
#define FOR(i,k,n) for(ll i=(k);i<(int)(n);i++)



vector<bool> gen_sosuu(vector<bool> sosuu){
  int size = sosuu.size();
  REP(i,size){
    sosuu[i] = true;
  }
  sosuu[0] = false;
  sosuu[1] = false;
  sosuu[2] = true;
  FOR(i,2,sqrt(size)+1){
    if(sosuu[i] == false) continue;
    for(int j = 2; i*j<size;j++){
      sosuu[i*j] =false;
    }
  }
  return sosuu;
}

int n;
ll a[100001];
ll b[100001];
ll c[100001];
ll mod = 1e9+7;

int main(){
  cin >> n;
  REP(i,n){
    cin >> a[i];
  }
  REP(i,n){
    cin >> b[i];
  }
  REP(i,n){
    c[i] = min(a[i],b[i]);
  }
  ll ans = 1;
  if(a[n-1]!=b[0]){
    cout << 0 <<endl;
    return 0;
  }
  int cnt = 2;
  FOR(i,1,n){
    if(cnt<2&&c[i]!=b[0]) break;
    if(c[i]==b[0]&&cnt>0){
     cnt--;
     continue;
    }
    if(c[i]==c[i-1]){
      ans = (ans*c[i]) %mod;
    }
    //cout << c[i] <<" " <<ans<<endl;
  }
  for(ll i = n-2;i>=0;i--){
    if(c[i]==b[0]) break;
    if(c[i]==c[i+1]){
      ans = (ans*c[i]) %mod;
    }
  }
  cout << ans <<endl;
}