#include<bits/stdc++.h>
using namespace std;

const int mn=4e5+6;
long long d[mn],f[mn][2];

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%lld",&d[i]);
	f[1][1]=-1e18;
	for(int i=2;i<=n;++i) {
		f[i][1]=max(f[i-2][1],f[i-2][0])+d[i];
		f[i][0]=f[i-2][0]+d[i-1];
		if(i&1) f[i][1]=max(f[i][1],max(f[i-3][1],f[i-3][0])+d[i]);
	}
	printf("%lld",max(f[n][0],f[n][1]));
}
### Prompt

Develop a solution in Cpp 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;

const int mn=4e5+6;
long long d[mn],f[mn][2];

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%lld",&d[i]);
	f[1][1]=-1e18;
	for(int i=2;i<=n;++i) {
		f[i][1]=max(f[i-2][1],f[i-2][0])+d[i];
		f[i][0]=f[i-2][0]+d[i-1];
		if(i&1) f[i][1]=max(f[i][1],max(f[i-3][1],f[i-3][0])+d[i]);
	}
	printf("%lld",max(f[n][0],f[n][1]));
}
```