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 lint;
typedef pair<int, int> pi;

int n, a[100005], b[100005];
int hs[100005], he[100005];
int main(){
	cin >> n;
	for(int i=1; i<=n; i++) scanf("%d",&a[i]);
	for(int i=1; i<=n; i++) scanf("%d",&b[i]);
	for(int i=1; i<=n; i++){
		if(a[i] != a[i-1]){
			hs[i] = he[i] = a[i];
		}
		else{
			hs[i] = 1;
			he[i] = a[i];
		}
	}
	for(int i=n; i; i--){
		if(b[i] != b[i+1]){
			if(max(hs[i], b[i]) > min(he[i], b[i])){
				puts("0");
				return 0;
			}
			hs[i] = he[i] = b[i];
		}
		else{
			he[i] = min(he[i], b[i]);
		}
	}
	lint ret = 1;
	for(int i=1; i<=n; i++){
		if(hs[i] > he[i]) ret = 0;
		ret *= he[i] - hs[i] + 1;
		ret %= 1000000007;
	}
	cout << ret << endl;
}
