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;

#define int long long
int dp[223456][5],v[223456];

int f(int S,int J){
  if(S<=0&&J-S==1)return 0;
  if(S<=0||J<0)return -1e18;
  if(~dp[S][J])return dp[S][J];
  return dp[S][J]=max({f(S-2,J),f(S-3,J-1),f(S-4,J-2)})+v[S-1];
}

signed main(){
  int n;cin>>n;
  for(int i=0;i<n;i++)cin>>v[i];
  memset(dp,-1,sizeof(dp));
  if(n&1)cout<<max({f(n,2),f(n-1,1),f(n-2,0)})<<endl;
  else cout<<max(f(n,1),f(n-1,0))<<endl;
}
