#include <bits/stdc++.h>
using namespace std;
int n,a[200005];
map<int,long long> dp[200005];
long long solve(int idx,int cnt)
{
	if (idx>=n)
	{
		if (cnt==n/2)
		return 0;
		return -1e18;
	}
	if (dp[idx].count(cnt))
	return dp[idx][cnt];
	if (cnt+(n-idx+1)/2<n/2)
	return -1e18;
	return dp[idx][cnt]=max(solve(idx+1,cnt),solve(idx+2,cnt+1)+a[idx]);
}
int main()
{
	scanf("%d",&n);
	for (int i=0;i<n;i++)
	scanf("%d",&a[i]);
	printf("%lld",solve(0,0));
}
### 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;
int n,a[200005];
map<int,long long> dp[200005];
long long solve(int idx,int cnt)
{
	if (idx>=n)
	{
		if (cnt==n/2)
		return 0;
		return -1e18;
	}
	if (dp[idx].count(cnt))
	return dp[idx][cnt];
	if (cnt+(n-idx+1)/2<n/2)
	return -1e18;
	return dp[idx][cnt]=max(solve(idx+1,cnt),solve(idx+2,cnt+1)+a[idx]);
}
int main()
{
	scanf("%d",&n);
	for (int i=0;i<n;i++)
	scanf("%d",&a[i]);
	printf("%lld",solve(0,0));
}
```