#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;
using ll = long long;

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){
        	dp[i+1][j+1] = max(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);
        }
    }
  
  	ll ans = dp[n][k];
  	cout << ans << endl;
}

### Prompt

Generate a cpp solution 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 P = pair<int, int>;
using ll = long long;

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){
        	dp[i+1][j+1] = max(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);
        }
    }
  
  	ll ans = dp[n][k];
  	cout << ans << endl;
}

```