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;
int main(){
	int n;
	cin>>n;

	ll a[n+1];
	for(int i =1;i <= n;i++){
		cin>>a[i];
	}

	ll ans[n+1];
	ans[1] = a[1];
	ans[2] = max(a[1],a[2]);

	if(n >= 3)
		ans[3] = max(a[2],a[3]);
	ll odd_sum[n+1];
	odd_sum[1] = a[1];

	for(int i =3;i <= n;i+=2)
		odd_sum[i] = odd_sum[i-2] + a[i];
	for(int i =4;i <= n;i++){
		if(i % 2 == 0)
			{ans[i] = ans[i-2] + a[i];
				ans[i] = max(ans[i],odd_sum[i-1]); //consider a;; odd indcices value
			}
		else
			ans[i] = max(ans[i-1],ans[i-2] + a[i]);
	}

	// for(int i =1;i<=n;i++)
	// 	cout<<ans[i]<<" ";
	// cout<<"\n";

	cout<<ans[n]<<"\n";

	return 0;
}