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: Accepted)

#include <bits/stdc++.h>
#define ALL(A) (A).begin(), (A).end()
#define ll long long
#define rep(i, n) for (int i = 0; i < (n); i++)

using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }

const ll mod = 1e9 + 7;
const ll INF = -1 * ((1LL << 63) + 1);
const int inf = -1 * ((1 << 31) + 1);

int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    int n;
    cin >> n;
    vector<int> a(n),b(n),c(n,-1);
    rep(i,n)cin >> a[i];
    rep(i,n)cin >> b[i];
    c[0] = a[0];
    c[n-1] = b[n-1];
    if(a[n-1] != b[0]){
        cout << 0 << endl;
        return 0;
    }
    for(int i=1;i<n;i++){
        if(a[i]>a[i-1]){
            if(c[i]!=-1 && c[i]!=a[i]){
                cout << 0 << endl;
                return 0;
            }else c[i] = a[i];
        }else if(c[i]!=-1 && c[i] >a[i]){
            cout << 0 << endl;
            return 0;
        }
    }
    for(int i=n-2;i>=0;i--){
        if(b[i]>b[i+1]){
            if(c[i]!=-1 && b[i]!=c[i]){
                cout << 0 << endl;
                return 0;
            }else c[i] = b[i];
        }else if(c[i]!=-1 && c[i]>b[i]){
            cout << 0 << endl;
            return 0;
        }
    }
    ll ans = 1;
    for(int i=0;i<n;i++){
        if(c[i]==-1){
            ans *= min(a[i],b[i]);
            ans %= mod;
        }
    }
    cout << ans << endl;
}