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 <cstdio>
#include <algorithm>
using namespace std;

const int MAXN = 100005;
const int MOD = 1000000007;

int N;
int t[MAXN], a[MAXN];
int x[MAXN], y[MAXN];

void load() {
	scanf("%d", &N);
	for (int i = 1; i <= N; i++)
		scanf("%d", t + i);
	for (int i = 1; i <= N; i++)
		scanf("%d", a + i);
}

int solve() {
	int sol = 1;
	for (int i = 1; i <= N; i++) {
		if (t[i] > t[i - 1])
			x[i] = t[i];
		if (a[i] > a[i + 1])
			y[i] = a[i];
		if (x[i] && y[i] && x[i] != y[i])
			return 0;
		if (y[i] > t[i] || x[i] > a[i])
			return 0;
		if (!x[i] && !y[i])
			sol = (long long)sol * min(t[i], a[i]) % MOD;
	}
	
	return sol;	
}

int main() {
	load();
	printf("%d\n", solve());
	return 0;
}