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>
using namespace std;
typedef long long ll;
const int maxn=2e5+50;
int t,n,m;
int arr[maxn];
ll dp[maxn];
int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>arr[i];
	}
/*  	if(n%2==0)
	{
		ll fin0=0;
		ll fin1=0;
		for(int i=1;i<=n;i++)
		{
			if(i&1) fin1+=arr[i];
			else fin0+=arr[i];
		}
		cout<<max(fin0,fin1)<<endl;
		return 0; 
	}*/
	ll sum[maxn];
	sum[1]=arr[1];
	for(int i=3;i<=n;i+=2)
	{
		sum[i]=sum[i-2]+arr[i];
	}
	for(int i=2;i<=n;i++)
	{
		if(i%2==1)
		{
			dp[i]=max(dp[i-1],dp[i-2]+arr[i]);
		}
		else
		{
			dp[i]=max(sum[i-1],dp[i-2]+arr[i]);
		}
	}
	cout<<dp[n]<<endl;
	return 0;
}