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;
#define _GLIBCXX_DEBUG //これつけるとA[N]でもいいらしい
//for文のマクロ
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define big 1000000007
#define all(a) sort((a).begin(),(a).end()) //ソートのマクロ
#define Re(a) reverse((a).begin(),(a).end())
#define YN(a) if(a){cout<<"yes"<<endl;}else cout<<"no"<<endl;//条件によってYes、Noを出力する
int main(){
    int n;
    cin>>n;
    vector<int> a(n),t(n);
    int amax=0,tmax=0;
    vector<int> l(n),r(n);
    rep(i,n){
        cin>>t[i];
        //tmax=max(tmax,t[i]);
    }
    rep(i,n){
        cin>>a[i];
        //amax = max(amax,a[i]);
    }
    if(n==1){
        if(a[0]==t[0]){
            cout<<1<<endl;
        }
        else{
            cout<<0<<endl;
        }
        return 0;
    }
    l[0]=1;
    for(int i=1;i<n;i++){
        if(t[i] == t[i-1]){
            l[i] = t[i];
        }
        else{
            l[i] = 1;
        }
    }
    r[n-1]=1;
    for(int i=n-2;i>=0;i--){
        if(a[i]==a[i+1]){
            r[i] = a[i];
        }
        else{
            r[i]=1;
        }
    }
    //どちらも上がるポイント=>l==r==1のときのaとtが一致してないとおかしい
    rep(i,n){
        if(l[i]==1 && r[i]==1 && a[i]!=t[i]){
            cout<<0<<endl;
            return 0;
        }
    }
    int64_t ans=1;
    rep(i,n){
        cout<<l[i]<<" ";
        ans = ans * min(r[i],l[i]) % big;
    }
    cout<<endl;
    rep(i,n){
        cout<<r[i]<<" ";
    }
    cout<<endl;
    cout<<ans<<endl;
}