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;
using ll=long long;
constexpr ll mod=1000000007;

int N,K;
ll A[200000],DP[222222][3];
int main(){
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>A[i];
	}
	for(int i=0;i<222222;i++){
		for(int j=0;j<3;j++){
			DP[i][j]=-1e18;
		}
	}
	DP[0][0]=DP[1][1]=DP[2][2]=0;
	for(int i=0;i<N;i++){
		for(int j=0;j<3;j++){
			DP[i][j]+=A[i];
			for(int k=0;j+k<3;k++){
				DP[i+2+k][j+k]=max(DP[i+2+k][j+k],DP[i][j]);
			}
		}
	}
	if(N&1){
		cout<<DP[N+1][2]<<endl;
	}
	else{
		cout<<DP[N+1][1]<<endl;
	}
}
