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 INF 2000000000
#define MOD 1000000007
#define EPS (1e-10)

using namespace std;


vector<int>  dx = {1,-1,0,0};
vector<int>  dy = {0,0,1,-1};


int main() {

	long n; cin >> n;

	vector<long> ans(n,0);
	vector<long> h(n,0);
	vector<long> l(n,0);

	long old_h = 0;
	vector<long> T(n);
	vector<long> A(n);
	for (int i = 0; i < n; i++) {
		cin >> T[i];	
	}
	for (int i = 0; i < n; i++) {
		cin >> A[i];	
	}

	for (int i = 0; i < n; i++) {
		h[i] = T[i];
		if (T[i] > old_h) {
			ans[i] = 1;
			l[i] = T[i];
		}
		old_h = T[i];
	}

	old_h = 0;
	for (int i = n-1; i >= 0; i--) {
		h[i] = min(A[i],h[i]);
		if (A[i] > old_h) {
			if (h[i] < A[i] || l[i] > A[i]) {
				cout << 0 << endl;
				return 0;
			}
			ans[i] = 1;
		}
		old_h = A[i];
	}

	

	long ANS = 1;
	for (int i = 0; i < n; i++) {
		if (ans[i] == 0) {
			ANS *= h[i] % MOD;	
			ANS %= MOD;
		}
	}

	if (T[n-1] != A[0]) {
		cout << 0 << endl;
	}else {
		cout << ANS << endl;
	}




	return 0;
}