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
#include<iostream>
#include<math.h>
using namespace std;
typedef long long ll;

ll n;
ll a[200005];
ll dp[200005];
ll lastsum[200005];
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	
	lastsum[1]=a[1];
	for(int i=3;i<=n;i+=2)
	{
		lastsum[i]=lastsum[i-2]+a[i];
	}
	
	for(int i=2;i<=n;i++)
	{
		if(i%2==0)
		{
			dp[i]=max(dp[i-2]+a[i],lastsum[i-1]);
		}
		else
		{
			dp[i]=max(dp[i-2]+a[i],dp[i-1]);
		}
	}
	
	cout<<dp[n];
	return 0;
}