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;
#define ll long long
const ll INF=1e18;
ll dp[200005][3], arr[200005];
int n;
ll f(int i, int skip){
	if(i>=n){
		if(skip<=1) return 0ll;
		return -INF;
	}
	if(dp[i][skip]!=-INF) return dp[i][skip];
	dp[i][skip]=f(i+2, skip)+arr[i];
	if(skip>0) dp[i][skip]=max(dp[i][skip], f(i+1, skip-1));
	return dp[i][skip];
}
int main(){
	cin>>n;
	for(int i=0; i<n; i++){
		cin>>arr[i];
		for(int j=0; j<3; j++)
			dp[i][j]=-INF;
	}
	ll ans=f(0, (n&1)?2:1);
	if(n&1){
		ll tmp=0ll, mn=INF;
		for(int i=0; i<n; i+=2){
			tmp+=arr[i];
			if(mn>arr[i]) mn-=arr[i];
		}
		tmp-=mn;
		ans=max(ans, tmp);
	}
	cout<<ans<<endl;
}