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 <iostream>

using namespace std;

int main()
{
  long N;
  cin >> N;

  long long A[200000];


   long i;

   for (i=0;i<N;i++)
      cin >> A[i];

  long long DP[200000];
  long long SUM_A[200000];
  SUM_A[0] = A[0]; // on somme les impairs
  for (i=2;i<N;i+=2)
    SUM_A[i] = SUM_A[i-2] + A[i];


  DP[0] = 0;
  DP[1] = max(A[0],A[1]);

  for (i=3;i<=N;i++) {
    if (i % 2) { // si c'est 7, alors A_7 + DP(5) ou DP(6)
      DP[i-1] = max(DP[i-2],A[i-1]+DP[i-3]);
    }
    else { // si c'est 8
      DP[i-1] = max(SUM_A[i-2],DP[i-3]+A[i-1]);
    }
  }


   cout << DP[N-1] << endl;
  return 0;
}
