#include <bits/stdc++.h>
using namespace std;
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int N;
	long long sum;
	cin>>N;
	vector<long long>a(N);
	for(int i=0;i<N;i++) cin>>a[i];
	vector<long long>dp(N+1);
	sum=a[0];
	for(int i=2;i<=N;i++)
		if(i%2==0) dp[i]=max(a[i-1]+dp[i-2],sum);
		else {
			dp[i]=max(a[i-1]+dp[i-2],dp[i-1]);
			sum+=a[i-1];
		}
	cout<<dp[N]<<"\n";
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int N;
	long long sum;
	cin>>N;
	vector<long long>a(N);
	for(int i=0;i<N;i++) cin>>a[i];
	vector<long long>dp(N+1);
	sum=a[0];
	for(int i=2;i<=N;i++)
		if(i%2==0) dp[i]=max(a[i-1]+dp[i-2],sum);
		else {
			dp[i]=max(a[i-1]+dp[i-2],dp[i-1]);
			sum+=a[i-1];
		}
	cout<<dp[N]<<"\n";
	return 0;
}
```