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>
#define int long long
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define F first
#define S second

using namespace std;
typedef pair<int,int> P;

const int MOD=1000000007;
int INF=100100100100100100;

int a[200010];
int dp[200010][3];
signed main(){
  int n;cin>>n;
  int m=n/2;
  rep(i,n)cin>>a[i];
  for(int i=1;i<=m;i++){
    rep(j,3){
      int ans=-INF;
      rep(k,j+1){
        ans=max(ans,dp[i-1][k]+a[2*(i-1)+k]);
      }
      dp[i][j]=ans;
    }
  }
        
  cout<<dp[m][n-(2*m-1)]<<endl;    
  
  return 0;
}