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 <map>
#include <cstdio>
#include <cstring>
#include <math.h>
#include <bitset>
#include <time.h>
#include <set>
#include <algorithm>
#define MODD 1000000007
#define ll long long
using namespace std;

int T[111111];
int A[111111];

int main() {
    int N;
    cin>>N;
    T[0]=0;
    for(int i=1;i<=N;i++) {
        cin>>T[i];
    }
    
    for(int i=1;i<=N;i++)
        cin>>A[i];
    A[N+1]=0;
    
    ll ans = 1;
    for(int i=1;i<=N;i++) {
        if (T[i]>T[i-1] || A[i]>A[i+1]) {
            int x = (T[i]>T[i-1] ? T[i] : A[i]);
            
            if (x > T[i] || x > A[i])
                ans = 0;
            
        } else {
            ans = (ans * min(A[i], T[i]))%MODD;
        }
    }
    
    cout << ans << endl;
    
}