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 long long long
using namespace std;
const int N=1e6;
long aa[N+2];
map<int,long>dp[N+2];
int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  int n;cin>>n;
  for(int i=1;i<=n;i++)
  {
    cin>>aa[i];
    //aa[i]=rand();
  }

  dp[0][0]=0;dp[1][0]=0;dp[1][1]=aa[1];
  for(int i=2;i<=n;i++)
  {
    for(int j=i/2-4;j<=i/2+4;j++)
    {
      if(j<0)continue;if(j==0)dp[i][j]=0;
      dp[i][j]=-1e18;
      if(dp[i-1].count(j))dp[i][j]=dp[i-1][j];
      if(dp[i-2].count(j-1))dp[i][j]=max(dp[i][j],aa[i]+dp[i-2][j-1]);
    }
  }
  cout<<dp[n][n/2]<<endl;
  return 0;
}