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)

//
//  cakes.cpp
//  Competitions
//
//  Created by macbook on 10/23/16.
//  Copyright © 2016 LiorCohen. All rights reserved.
//

#include <iostream>
#include <vector>

using namespace std;

int main() {
    
    std::ios_base::sync_with_stdio(false);  //make input/output faster!!!
    
    unsigned int N;
    cin >> N;
    
    vector<unsigned int> Tak(N),Aoki(N),Sure(N,0);
    
    for (int i = 0; i<N; i++) {
        cin >> Tak[i];
        if (i == 0) {
            Sure[i] = Tak[i];
        } else {
            if (Tak[i] > Tak[i-1]) {
                Sure[i] = Tak[i];
            }
            if (Tak[i] == 1) {
                Sure[i] = 1;
            }
        }
    }
    for (int i = 0; i<N; i++) {
        cin >> Aoki[i];
    }
    
    //traverse aoki from right to left:
    for (int i=N-1; i>=0; i--) {
        if (i == N-1) {
            //  this is sure
            if ((Aoki[i] > Tak[i]) || ( Sure[i] != 0 && Sure[i] != Aoki[i] )) {
                // illegal
                cout << 0 << endl;
                return 0;
            }
            Sure[i] = Aoki[i];
        }
        if (Aoki[i] == 1 || Aoki[i] > Aoki[i+1]) {
            if (( Sure[i] != 0 && Sure[i] != Aoki[i] )) {
                // illegal
                cout << 0 << endl;
                return 0;
            }
            Sure[i] = Aoki[i];
        }
        
    }
    
    unsigned int psbl = 1;
    for (int i=0; i<N; i++) {
        if (Sure[i] == 0) {
            unsigned int curPsbl = (Aoki[i] < Tak[i]) ? Aoki[i] : Tak[i];
            psbl = psbl * curPsbl;
        }
    }
    
    cout << psbl << endl;
    
    return 0;
}