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 < (int)(n); i++)

int64_t dp[4][2];

int main(){
  int n;
  cin >> n;
  vector<int> a(n);
  REP(i, n) cin >> a[i];
  REP(i, n){
    if(i % 2 == 0){
      dp[0][1] = dp[1][0] + a[i];
      dp[2][1] = dp[3][0] + a[i];
      dp[2][0] = max(dp[1][0], dp[1][1]);
    }else{
      dp[1][1] = dp[2][0] + a[i];
      dp[3][0] = max(dp[2][0], dp[2][1]);
      dp[1][0] = dp[0][1];
    }
  }
  if(n % 2 == 0){
    cout << max(dp[1][0], dp[1][1]) << endl;
  }else{
    cout << max(dp[2][0], dp[2][1]) << endl;
  }
  return 0;
}