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>

using namespace std;

long long en[200000],me[200000][2];
int vi[200000][2];
int n;

long long dp(int ac, int v){
	if(ac<0)return 0;
	if(vi[ac][v])return me[ac][v];
	vi[ac][v]=1;
	if(v)return me[ac][v] = en[ac] + dp(ac-2,v);
	if(ac%2){
		return me[ac][v] = max(dp(ac-1,1), en[ac] + dp(ac-2,v));
	}
	if(ac==0)return me[ac][v]=0;
	
	return me[ac][v] = max(dp(ac-1,v),en[ac] + dp(ac-2,v));
}
int main(){
	
	cin.tie(0); ios_base::sync_with_stdio(0);
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>en[i];
	}
	cout<<dp(n-1,0);
	
	return 0;
}