#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
#define chmax(x,y) x = max(x,y)
 
const ll INF = 1e18;
 
ll dp[200005][4];
 
int main() {
  int n;
  cin>>n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  int k = 1+n%2;
  rep(i,n+1)rep(j,k+1) dp[i][j] = -INF;
  dp[0][0] = 0;
  rep(i,n) {
    rep(j,k+1) {
      chmax(dp[i+1][j+1], dp[i][j]);
      ll now = dp[i][j];
      if ((i+j)%2 == 0) now += a[i];
      chmax(dp[i+1][j], now);
    }
  }
  ll ans = dp[n][k];
  cout << ans << endl;
  return 0;
}
### Prompt

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
#define chmax(x,y) x = max(x,y)
 
const ll INF = 1e18;
 
ll dp[200005][4];
 
int main() {
  int n;
  cin>>n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  int k = 1+n%2;
  rep(i,n+1)rep(j,k+1) dp[i][j] = -INF;
  dp[0][0] = 0;
  rep(i,n) {
    rep(j,k+1) {
      chmax(dp[i+1][j+1], dp[i][j]);
      ll now = dp[i][j];
      if ((i+j)%2 == 0) now += a[i];
      chmax(dp[i+1][j], now);
    }
  }
  ll ans = dp[n][k];
  cout << ans << endl;
  return 0;
}
```