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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll p = 1e9+7;
const int MAXN = 1e5+5;

ll prod = 1;
ll l[MAXN], r[MAXN], n;

int main() {
	cin >> n;
	for (int i = 1; i <= n; ++i)
		cin >> l[i];
	for (int i = 1; i <= n; ++i)
		cin >> r[i];
	for (int i = 1; i <= n; ++i) {
		if (l[i] < l[i-1] || r[i] < r[i+1]) {
			cout << "0\n"; return 0;
		} if (l[i] != l[i-1] && r[i] != r[i+1]) {
			if (l[i] != r[i]) {
				cout << "0\n"; return 0;
			} else
				continue;
		} if (l[i] != l[i-1]) {
			if (l[i] > r[i]) {
				cout << "0\n"; return 0;
			} continue;
		} if (r[i] != r[i+1]) {
			if (r[i] > l[i]) {
				cout << "0\n"; return 0;
			} continue;
		}

		prod = prod*min(l[i], r[i])%p;
	}

	cout << prod << endl;
}