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>
long long int n, a[100010], b[100010];
int main(){
    long long int s=1,i,max=0,max1=0,y1=0,y2=0;
    scanf("%lld",&n);
    for(i=0;i<n;i++){
        scanf("%lld",a+i);
        if(max<a[i])max=a[i],y1=i;
    }
    for(i=0;i<n;i++){
        scanf("%lld",b+i);
        if(max1<=b[i])max1=b[i],y2=i;
        
    }
    if(max!=max1||y1>y2){
        printf("0");
        return 0;
    }
    max=0;
    for(i=0;i<=y1;i++){
        if(max>=a[i]){
            s=(s*a[i])%1000000007;
        }else{
            max=a[i];
        }
    }
    for(i=y1+1;i<y2;i++){
        s=(s*a[i])%1000000007;
    }
    max=0;
    for(i=n-1;i>=y2;i--){
        if(max>=b[i]){
            s=(s*b[i])%1000000007;
        }else{
            max=b[i];
        }
    }
    printf("%lld",s);
}