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;

const ll INF = 1e15;

int main() {
  int N, K;
  cin >> N;
  K = 1 + N % 2;
  vector<ll> A(N);
  for (int i = 0; i < N; i++) cin >> A[i];

  vector<vector<ll>> dp(N + 1, vector<ll>(K + 2, -INF));

  dp[0][0] = 0;
  for (int i = 0; i <= N - 1; i++) {
    for (int j = 0; j <= K; j++) {
      dp[i + 1][j + 1] = dp[i][j];
      ll now = dp[i][j];
      if ((i + j) % 2 == 0) now += A[i];
      dp[i + 1][j] = max(dp[i + 1][j], now);
    }
  }
  cout << dp[N][K] << endl;
}
