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>
#define MAXN 100005
using namespace std;
const int mod=1e9+7;
int t[MAXN],a[MAXN],n;
int mul(int x,int y) 
{
    return 1LL * x * y % mod;
}
int main() 
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%d",&t[i]);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    int ans=1;
    for(int i=1;i<=n;++i) 
	{
	    if(t[i]!=t[i-1]||a[i]!=a[i + 1]) 
		{
	        int x=-1,y=-1;
	        if(t[i]!=t[i-1]) x=t[i];
	        if(a[i]!=a[i+1]) y=a[i];
	        if(x!=-1&&y!=-1) 
		        if(x!=y) 
		        {
		        	printf("0");
					return 0;
				}		
	        if(y>t[i]) 
			{
				printf("0");
				return 0;
			}
	        if(x>a[i]) 
			{
				printf("0");;
				return 0;
			}
	    }
	    else 
	        ans=mul(ans,min(a[i],t[i]));
    }
    printf("%d",ans);
    return 0;
}