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 n;
long long a[200003];
long long dp[200003];
long long sumb[200003];
long long ans=-2e18;
long long sum;
long long su; 
int main(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	dp[1]=a[1];
	dp[2]=a[2];
	sumb[n]=a[n];
	sumb[n-1]=a[n-1];
	for(int i=3;i<=n;i++) dp[i]=dp[i-2]+a[i];
	for(int i=n-2;i>=1;i--) sumb[i]=sumb[i+2]+a[i];
	sum=dp[n];
	if(n%2==0){
		ans=max(dp[n-1],dp[n]);
		for(int i=1;i<=n;i+=2) ans=max(ans,dp[i]+sumb[i+3]);
		cout<<ans<<endl;
		return 0;
	}
	for(int i=1;i<=n;i++){
		if(i%2==0) su+=a[i];
		else{
			su=max(su-a[i],-a[i]);
			ans=max(ans,sum+su);
		}
	}
	cout<<ans<<endl;
	return 0;
}