Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

* 2 \leq N \leq 2\times 10^5
* |A_i|\leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 ... A_N


Output

Print the maximum possible sum of the chosen elements.

Examples

Input

6
1 2 3 4 5 6


Output

12


Input

5
-1000 -100 -10 0 10


Output

0


Input

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000


Output

5000000000


Input

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49


Output

295
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;
  cin>>N;
  vector<int64_t>A(N);
  for(int i=0;i<N;i++)
    cin>>A.at(i);
  vector<int64_t>B(1+N);
  map<int64_t,int64_t>M;
  M[1]=A.at(0);
  for(int i=2;i<=N;i++)
    if(i%2==1)
      M[i]=M.at(i-2)+A.at(i-1);
  B.at(0)=0;
  B.at(0)=0;
  B.at(2)=max(A.at(0),A.at(1));
  for(int i=3;i<=N;i++){
  if(i%2==0)
    B.at(i)=max(A.at(i-1)+B.at(i-2),A.at(i-2)+M.at(i-3));
    else
      B.at(i)=max(A.at(i-1)+B.at(i-2),A.at(i-2)+B.at(i-3));
  
  }cout<<B.at(N)<<endl;
   return 0;
}