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<bits/stdc++.h>
#define int long long
using namespace std;
int n,a[200003],fsum[200003],bsum[200003],ans=-2e18,sum,su;
signed main()
{
	scanf("%lld",&n);
	for(int i=1; i<=n; i++)
		scanf("%lld",&a[i]);
	fsum[1]=a[1],fsum[2]=a[2],bsum[n]=a[n],bsum[n-1]=a[n-1];
	for(int i=3; i<=n; i++)
		fsum[i]=fsum[i-2]+a[i];
	for(int i=n-2; i>=1; i--)
		bsum[i]=bsum[i+2]+a[i];
	sum=fsum[n];
	if(n%2==0)
	{
		ans=max(fsum[n-1],fsum[n]);
		for(int i=1; i<=n; i+=2)
			ans=max(ans,fsum[i]+bsum[i+3]);
		cout<<ans;
		return 0;
	}
	for(int i=1; i<=n; i++)
		if(i%2==0)
			su+=a[i];
		else
			su=max(su-a[i],-a[i]),ans=max(ans,sum+su);
	cout<<ans;
	return 0;
}
