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 <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#define M 1000000007
using namespace std;

long t[1000000] = {0};
long a[1000000] = {0};
bool tr[1000000] = {0};
bool ar[1000000] = {0};

int fail() {
    cout << 0 << endl;
    return 0;
}

int main() {
    int n;
    long mt = 0, ma = 0;
    vector< pair<long, int> > res;
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> t[i];
        if(t[i] > mt) mt = t[i];
        if(i != 0 && t[i] == t[i - 1]) tr[i] = 1;
//        cout << int(tr[i]) << " ";
    }
//    cout << endl;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        if(a[i] > ma) ma = a[i];
        if(i != 0 && a[i] == a[i - 1]) ar[i - 1] = 1;
    }
    if(mt != ma) return fail();
    int cnt = 0;
    long val = 0;
    for(int i = 0; i < n; i++) {
        if(tr[i] && ar[i]) {
            if(cnt == 0) val = min(t[i], a[i]);
            cnt++;
        } else if(cnt != 0) {
            res.push_back(make_pair(val, cnt));
            cnt = 0;
            val = 0;
        } else if( (t[i] > a[i] && ar[i]) || (t[i] < a[i] && tr[i]) ){
            return fail();
        }
    }
    if(res.size() == 0) {
        cout << 1 << endl;
    } else {
        long r = 1;
        for(auto it = res.begin(); it != res.end(); it++) {
            for(int i = 0; i < it->second; i++) {
                r *= it->first;
                r %= M;
            }
        }
        cout << r << endl;
    }
    return 0;
}