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>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> P;
#define pb push_back
#define mp make_pair
#define fr(i,n) for(int i=0;i<n;i++)
#define Fr(i,n) for(int i=0;i++<n;)
#define ifr(i,n) for(int i=n-1;i>=0;i--)
#define iFr(i,n) for(int i=n;i>0;i--)

const ll MOD=1e9+7;

ll ad(ll x,ll y=0){
  return (x+y)%MOD;
}

ll mlt(ll x,ll y=1){
  return (x*y)%MOD;
}

int main(){
  ll n,t[100010],a[100010],m[100010]={},M[100010]={},ans=1;
  cin>>n;
  Fr(i,n) cin>>t[i];
  fr(i,n) cin>>a[i];
  t[0]=0;a[n]=0;
  Fr(i,n){
    M[i]=t[i];
    if(t[i]!=t[i-1]) m[i]=t[i];
    else m[i]=1;
  }
  iFr(i,n){
    M[i]=min(M[i],a[i-1]);
    if(a[i]!=a[i-1]) m[i]=max(m[i],a[i-1]);
  }
  Fr(i,n) ans=mlt(ans,max(0ll,M[i]-m[i]+1));
  cout<<ans<<endl;
}