#include "bits/stdc++.h"

using namespace std;

typedef long long ll;
const ll N = 2e5+5;

ll a[N], d[N], s[N], n, i;

int main(){
	cin >> n;
	for(; i < n; ++i)
	{
		cin >> a[i];
		s[i] = (i > 1 ? s[i - 2] : 0) + a[i];
	}
	for (i = 1; i < n; ++i)
	{
		if (i & 1)
			d[i] = max(s[i - 1], d[i - 2] + a[i]);
		else
			d[i] = max(d[i - 1], d[i - 2] + a[i]);
	}
	cout << d[n - 1];
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;

typedef long long ll;
const ll N = 2e5+5;

ll a[N], d[N], s[N], n, i;

int main(){
	cin >> n;
	for(; i < n; ++i)
	{
		cin >> a[i];
		s[i] = (i > 1 ? s[i - 2] : 0) + a[i];
	}
	for (i = 1; i < n; ++i)
	{
		if (i & 1)
			d[i] = max(s[i - 1], d[i - 2] + a[i]);
		else
			d[i] = max(d[i - 1], d[i - 2] + a[i]);
	}
	cout << d[n - 1];
}
```