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 <functional>
#include <string>
#include <vector>
#include <math.h>

using namespace std;

int main(void) {
	int n;
	
	cin >> n;
	vector<unsigned long int>a(n);
	vector<unsigned long int>b(n);
	vector<unsigned long int>t(n, 0);
	
	for (long int i = 0; i < n; i++)
	{
		cin >> a[i];
		if (i == 0)
		{
			t[i] = a[i];
		}
		else
		{
			if (a[i] > a[i - 1])
				t[i] = a[i];
		}
	}

	int flag = 0;

	for (long int i = 0; i < n; i++)
	{
		cin >> b[i];

		if (i == n - 1)
		{
			if (t[i] == 0)
				t[i] = b[i];
			else if (t[i] != b[i])
				flag = -1;
		}
		if (i > 0)
		{
			if (b[i - 1] > b[i])
			{
				if (t[i - 1] == 0)
					t[i -1] = b[i-1];
				else if (t[i-1] != b[i-1])
					flag = -1;
			}
		}
	}


	unsigned long int ret = 1;

	unsigned long int bounder1;
	unsigned long int bounder2;

	if (flag >= 0)
	{
		int f = 0;
		long int left, right;
		for (long int i = 0; i < n; i++)
		{
		
			if ((f == 1) && (t[i] != 0))
			{
				bounder2 = t[i];
				right = i;

				unsigned long int bounder = min(bounder1, bounder2);
				long int k = right - left;
				long long int l = 1;
				long int ten = pow(10, 9) + 7;
				for (long int j = 0; j < k; j++)
				{
					long long int ltemp = l* bounder;
					l = ltemp % ten;
				}
				long long int tempret = ret * (l % ten);
				ret = tempret % ten;

				f = 0;
			}
			if ((f == 0) && (t[i] == 0))
			{
				bounder1 = t[i - 1];
				left = i;
				f = 1;
			}
		}
	}
	else
		ret = 0;

	
	cout << ret << endl;

	return 0;
}