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()
{
    long long int j,n,m,po,no,flag,cnt=0,cnt2,ar[200001],x,y,i,co[200001],sum=0,k,ans=INT_MIN,sum1=0;
    string str[1000],ptr,sr,tr,rtr,rt;
    vector<long long int>v1,v2;
    set<string>s;
    queue<long long int>qu;
    set<string>::iterator it;
    cin>>n;
    for(i=1; i<=n; i++)
    {
        cin>>ar[i];

    }
    co[0]=0;
    co[1]=0;
    x=ar[1];
   for(i=2;i<=n;i++)
   {
       if(i%2==0)
        co[i]=max((co[i-2]+ar[i]),x);
        else
       {
           co[i]=max((co[i-2]+ar[i]),co[i-1]);
           x+=ar[i];
       }
   }



    cout<<co[n]<<endl;



}
