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>
#include<algorithm>
using namespace std;
int N;
long A[2<<17];
long dp[2<<17][3];
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)cin>>A[i];
	for(int i=0;i<=N;i++)for(int j=0;j<3;j++)dp[i][j]=-9e18;
	dp[0][0]=0;
	for(int i=0;i<N;i++)for(int j=0;j<3;j++)
	{
		if(dp[i][j]<-1e18)continue;
		dp[min(i+2,N)][j+(i+2==N)]=max(dp[min(i+2,N)][j+(i+2==N)],dp[i][j]+A[i]);
		if(j<2)dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]);
	}
	cout<<dp[N][N%2?2:1]<<endl;
}