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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int mod = 1000000007;
class mint {
    long long x;
public:
    mint(long long x=0) : x((x%mod+mod)%mod) {}
    mint operator-() const { 
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    mint choose(int n, int a){
        mint x = 1 ,y = 1;
        for(int i=0; i<a; ++i){
            x *= n-i;
            y *= i+1;
        }
        return x / y;
    }
    friend istream& operator>>(istream& is, mint& m){
        is >> m.x;
        return is;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
};

const int INF=1001001001;

int main(){
    int n;
    cin>>n;
    vector<int>t(n),a(n),h(n,-1);
    rep(i,n)cin>>t[i];
    rep(i,n)cin>>a[i];
    if(n==1&&t[0]!=a[n-1]){cout<<0<<endl; return 0;}
    h[0]=t[0];
    h[n-1]=a[n-1];
    for(int i=1; i<n; ++i){
        if(t[i]>t[i-1]){
            if(h[i]==-1||h[i]==t[i])h[i]=t[i];
            else {cout<<0<<endl; return 0;}
        }
    }
    for(int i=n-2; i>=0; --i){
        if(a[i]>a[i+1]){
            if(h[i]==-1||h[i]==a[i])h[i]=a[i];
            else {cout<<0<<endl; return 0;}
        }
    }
    vector<int>h2(n);
    rep(i,n)h2[i]=min(a[i],t[i]);
    mint ans=1;
    rep(i,n){
        if(h[i]==-1)ans*=h2[i];
    }
    cout<<ans<<endl;
}