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 rep(i,n) for (int i = 0; i < (n); ++i)
using ll = long long;
using P = std::pair<int,int>;

const ll N = 2e5 + 5;
ll n,a[N],sum[N],f[N];

int main(){
   cin >> n;
   cin >> a[1] >> a[2];
   sum[1] = a[1];
   sum[2] = a[2];
   
   for(int i=3;i<=n;i++){
      cin >> a[i];
      sum[i] = sum[i-2] + a[i]; 
   }

   for(int i=2;i<=N;i++){
      if(i&1){
         f[i] = max(f[i-2] + a[i], f[i-1]);
      }else{
         f[i] = max(f[i-2] + a[i], sum[i-1]);
      } 
   }

   cout << f[n] << endl;
   return 0;  
}