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;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

int main() {
  ll n;
  cin >> n;
  vec a(n);
  for(ll i=0;i<n;i++) cin >> a[i];
  ll check=1;
  check+=n%2;
  mat dp(n+1,vec(check+1,-inf));
  dp[0][0]=0;
  for(ll i=0;i<n;i++) {
    for(ll j=0;j<=check;j++) {
      if(j!=check) {
        dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]);
      }
      ll cur=dp[i][j];
      if((i+j)%2==0) cur+=a[i];
      dp[i+1][j]=max(dp[i+1][j],cur);
    }
  }
  cout << dp[n][check] << endl;
}