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>
using namespace std;

typedef long long ll;
int N;
ll T[100010]={0},A[100010]={0},R[100010];
ll inf = 1e9+7;

int main(){
	cin >> N;
	for(int i=1;i<=N;i++){
		cin >> T[i];
	}
	for(int i=1;i<=N;i++){
		cin >> A[i];
	}
	R[1] = 1;
	R[N] = 1;
	bool judge = true;
	ll ans = 1;
	if(T[N]!=A[1]) judge = false;
	else{
		for(int i=1;i<=N;i++){
			if(T[i]>T[i-1] || A[i]>A[i+1]){
				R[i] = 1;
				if(T[i]>T[i-1] && (T[i]>A[i+1] && T[i]!=A[i])) judge = false;
				if(A[i]>A[i+1] && (A[i]>T[i-1] && T[i]!=A[i])) judge = false;
			}
			else R[i] = min(A[i],T[i]);
			ans = (ans*R[i])%inf;
		}
	}
	if(!judge) cout << 0 << endl;
	else cout << ans << endl;
}