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>
#include <vector>
using namespace std;

int main(){
  int N;
  cin >> N;
  vector<long long> A(N);
  for(int i = 0; i < N; ++i){
    cin >> A[i];
  }
  int m = 2+N%2;
  const long long INF = 1e18;
  vector<long long> dp(m,-INF);
  dp[0] = 0;
  for(int i = 0; i < N; ++i){
    vector<long long> dp_(m,-INF);
    for(int j = 0; j < m; ++j){
      if(dp[j] == -INF) continue;
      dp_[j] = max(dp_[j],dp[j]+((i+j)%2 ? 0 : A[i]));
      if(j+1 < m)
        dp_[j+1] = max(dp_[j+1],dp[j]);
    }
    swap(dp,dp_);
  }
  cout << dp.back() << endl;
}
