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<stdio.h>
#include<algorithm>
using namespace std;
long long mod=1000000007;
int b[110000];
int c[110000];
int R[110000];
int L[110000];
int main(){
	int a;scanf("%d",&a);
	for(int i=0;i<a;i++)scanf("%d",b+i);
	for(int i=0;i<a;i++)scanf("%d",c+i);
	long long ret=1;
	for(int i=0;i<a;i++)L[i]=1;
	for(int i=0;i<a;i++)R[i]=mod;
	for(int i=0;i<a;i++){
		if(i==0||b[i]>b[i-1]){
			L[i]=R[i]=b[i];
		}else{
			R[i]=b[i];
		}
	}
	for(int i=a-1;i>=0;i--){
		if(i==a-1||c[i]>c[i+1]){
			L[i]=max(L[i],c[i]);
			R[i]=min(R[i],c[i]);
		}else{
			R[i]=min(R[i],c[i]);
		}
	}
//	for(int i=0;i<a;i++)printf("%d %d\n",L[i],R[i]);
	for(int i=0;i<a;i++){
		if(L[i]>R[i])ret=0;
		else ret=ret*(R[i]-L[i]+1)%mod;
	}printf("%lld\n",ret);
}