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>
#define mod 1000000007
using namespace std;
typedef long long ll;
int n;
ll T[100001],A[100001];

ll solve(){
  ll mxT=-1,idxT=0;
  for(int i=0;i<n;i++) if(mxT<T[i])mxT=T[i],idxT=i;
  ll mxA=-1,idxA=0;
  for(int i=n-1;i>=0;i--) if(mxA<A[i])mxA=A[i],idxA=i;
  if(mxA!=mxT)return 0;
  ll ans=1;
  for(int i=1;i<n;i++){
    if(T[i]!=T[i-1]&&T[i]>A[i])return 0;
    if(A[i]!=A[i+1]&&A[i]>T[i])return 0;
    if(T[i]!=T[i-1])continue;
    if(A[i]!=A[i+1])continue;
    ans=ans*min(T[i],A[i])%mod;
  } 
    
  return ans;
}

int main(){
  cin>>n;
  for(int i=0;i<n;i++) cin>>T[i];
  for(int j=0;j<n;j++) cin>>A[j];
 
  cout<<solve()<<endl;

  return 0;
}
