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;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long ll;
ll N;
ll T[100001];
ll A[100001];
int main(){
	cin>>N;
	REP(i,N){
	    cin>>T[i];
	}
	REP(i,N){
	   cin>>A[i];
	}
	if(A[0]!=T[N-1]){
	   cout<<0<<endl;
	   return 0;
	}
	
	ll p=1;
	int b=0;
	REP(i,N-1){
	    if(i==0)continue;
	    if((A[i]-A[i-1]!=0)&&(T[i]-T[i-1]!=0)){
	        p*=(A[i]==T[i]);
	        break;
	    }
	    if(T[i]-T[i-1]!=0){
	        p*=T[i]<=A[i];
	       continue;
	    }
	    if(A[i]-A[i+1]!=0){
	       p*=A[i]<=T[i];
	       continue;
	    }
	    ll a=min(A[i],T[i]);
	    p=(p*a)%1000000007;
	}
        cout<<p%1000000007<<endl;
	return(0);
}