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 <string>
#include <vector>
#include <sstream>
#include <algorithm>
#include <cmath>
#include <functional>

using namespace std;

#define M 1000000007

int main() {
	long long int n;
	cin >> n;
	vector<long long int> t(n);
	vector<long long int> a(n);
	for (long long int i = 0; i < n; i++) {
		cin >> t[i];
	}
	for (long long int i = 0; i < n; i++) {
		cin >> a[i];
	}
	vector<long long int> c(n);
	vector<long long int> d(n);
	c[0] = t[0];
	for (long long int i = 1; i < n; i++) {
		if (t[i-1] < t[i]) {
			c[i] = t[i];
		} else {
			c[i] = -1;
		}
	}
	d[n-1] = a[n-1];
	for (long long int i = n-2; i >= 0; i--) {
		if (a[i+1] < a[i]) {
			d[i] = a[i];
		} else {
			d[i] = -1;
		}
	}
	bool h = true;
	vector<long long int> e(n);
	for (long long int i = 0; i < n; i++) {
		e[i] = max(c[i], d[i]);
		if (c[i] > 0 && d[i] > 0 && c[i] != d[i]) {
			h = false;
			break;
		}
	}

	if (h) {
		long long int st = e[0];
		long long int ed = e[n-1];
		long long int rr = 0;
		long long int sm = 1;
		for (long long int i = 1; i < n; i++) {
			rr = i;
			if (e[i] == -1) {
				for (long long int j = i+1; j < n; j++) {
					if (e[j] == -1) {
					} else {
						rr = j;
						ed = e[j];
						break;
					}
				}
				for (long long int j = i+1; j < rr; j++) {
					sm *= abs(st-ed);
					sm %= M;
				}
				i = rr-1;
			} else {
				st = e[i];
			}
		}
		cout << sm << endl;
	} else {
		cout << 0 << endl;
	}
	/*
	for (long long int i = 0; i < n; i++) {
		cout << c[i] << ' ' << d[i] << endl;
	}*/
	return 0;
}