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.

Examples

Input

5
1 3 3 3 3
3 3 2 2 2


Output

4


Input

5
1 1 1 2 2
3 2 1 1 1


Output

0


Input

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5


Output

884111967


Input

1
17
17


Output

1
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
long long n, a[100002], b[100002], l[100002], r[100002], p[100002], q[100002], sum = 1;
int main() {
	cin >> n;
	for (int i = 0; i < n; i++) { cin >> a[i]; }
	for (int i = 0; i < n; i++)cin >> b[i];
	for (int i = 0; i < n; i++) { l[i] = 1; r[i] = a[i]; p[i] = 1; q[i] = b[i]; }
	l[0] = a[0]; r[0] = a[0]; p[n - 1] = b[n - 1]; q[n - 1] = b[n - 1];

	bool OK = true;
	for (int i = 1; i < n; i++) {
		if (a[i - 1] != a[i]) { l[i] = a[i]; r[i] = a[i]; }
	}
	for (int i = n - 2; i >= 0; i--) {
		if (b[i + 1] != b[i]) { p[i] = b[i]; q[i] = b[i]; }
	}
	for (int i = 0; i < n; i++) {
		int C1 = max(l[i], p[i]);
		int C2 = min(r[i], q[i]);
		if (C1 > C2)sum = 0;
		else { sum *= (C2 - C1 + 1); }
		sum %= 1000000007;
	}
	cout << sum << endl;
	return 0;
}