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;
const int N = 1e5 + 5, MOD = 1e9 + 7;
int ex1[N], ex2[N], les1[N], les2[N], a[N], b[N];
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++ i) {
		scanf("%d", &a[i]);
		les1[i] = a[i];
		if (a[i] > a[i - 1]) ex1[i] = a[i];
	}
	for (int i = 1; i <= n; ++ i) scanf("%d", &b[i]);
	for (int i = n; i >= 1; -- i) {
		les2[i] = b[i];
		if (b[i] > b[i + 1]) ex2[i] = b[i];
	}
	int ans = 1;
	for (int i = 1; i <= n; ++ i) {
		if (ex1[i] && ex2[i] && ex1[i] != ex2[i]) {
			ans = 0;
			break;
		}
		if (!ex1[i] && !ex2[i]) ans = 1LL * ans * min(les1[i], les2[i]) % MOD; else {
			if (max(ex1[i], ex2[i]) > min(les1[i], les2[i])) {
				ans = 0;
				break;
			}
		}
	}
	printf("%d\n", ans);
}

