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>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
    int N;
    cin >> N;
    vector<vector<int> > vec(N,vector<int>(2));
    for(int i=0; i<N; i++){
        vec[i] = {1,1000000000};
    }
    int tmp=0;
    for(int i=0; i<N; i++){
        int T;
        cin >> T;

        if (T==tmp){
            vec[i][1]=T;
        }else{
            vec[i]={T,T};
        }
        tmp=T;
    }
    tmp=0;
    vector<int> arr(N,0);
    for(int i=0; i<N; i++){
        cin >> arr[i];
    }
    for(int i=N-1; i>=0; i--){
        int A=arr[i];

        if (A==tmp){
            vec[i][1]=min(vec[i][1],A);
        }else{
            vec[i]={max(A,vec[i][0]),min(A,vec[i][1])};
        }
        tmp=A;
    }
    long long ans=1;
    for(int i=0; i<N; i++){
        ans*=max(0,vec[i][1]-vec[i][0]+1);
        ans%=1000000007;
    }
    cout << ans << endl;
}