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;
const int MOD=1000000007;
void cntr(){ cout << 0 << "\n"; }
int main(){
int N; cin >> N; int L=N-1;
unsigned long long T[N], A[N], ans=1;
for(int i=0; i<N; i++){ cin >> T[i]; }  
for(int i=0; i<N; i++){ cin >> A[i]; }  
if(N==1){ if(A[0]==T[0]){ cout << 1 << "\n"; }else{ cntr(); } }
else if(A[0]<T[0] || (A[0]>A[1] && A[0]!=T[0])){ cntr(); }
else if(A[N-1]>T[N-1] || (T[N-1]>T[N-2] && A[N-1]!=T[N-1])){ cntr(); }
else{  
for(int i=1; i<N-1; i++){ 
  if(T[i-1]==T[i] && A[i]==A[i+1]){ ans=(ans*min(T[i], A[i]))%MOD; }
  else if(T[i-1]<T[i] && A[i]>A[i+1]){ if(A[i]!=T[i]){cntr(); return 0;} }
  else if(T[i-1]<T[i]){ if(A[i]<T[i]){cntr(); return 0;} }
  else if(A[i]>A[i+1]){ if(A[i]>T[i]){cntr(); return 0;} }
}
cout << ans << "\n";
}
}