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;
using ll = long long;
#define rep(i,n) for(int i=0;i<(n);++i)
#define chmax(x,y) x=max(x,y)
const ll INF = 1e18;
ll dp[200001][4]; // [N+1][skip+1]

int main() {
  int N; cin>>N;
  vector<int> A(N);
  rep(i, N) cin>>A[i];

  rep(i, N+1) rep(j, 4) dp[i][j] =-INF;
  dp[0][0] = 0;

  rep(i, N){
		rep(k, 3){
			chmax(dp[i+1][k], dp[i][k] + ((i+k)%2? 0:A[i]));
			chmax(dp[i+1][k+1], dp[i][k]);
		}
  }

  cout << dp[N][N%2+1] << endl;

}
