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++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,int> P;

int main() { 
  int n;
  cin >> n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];

  vector<ll> dp(n + 1, 0);
  ll sum = a[0];
  for (int i = 2; i <= n; i++) {
    if (i % 2 == 0) {
      dp[i] = max(a[i - 1] + dp[i - 2], sum);
    } else {
      dp[i] = max(a[i - 1] + dp[i - 2], dp[i - 1]);
      sum += a[i - 1];
    }
  }
  cout << dp[n] << endl;
  return 0;
} 