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;
typedef long long ll;
const ll N=2e5+7;
ll dp[N][2];
ll arr[N];
int main(){
	ll n;
	cin>>n;
	for(ll i=1;i<=n;i++)  cin>>arr[i];

	dp[2][0]=arr[1];
	dp[2][1]=arr[2];
	for(ll i=3;i<=n;i++){
		if(i&1){
			dp[i][1]=max({dp[i-2][1],dp[i-2][0],dp[i-3][1],dp[i-3][0]})+arr[i];
			dp[i][0]=max({dp[i-1][0],dp[i-1][1],arr[i-2]+dp[i-3][0]});
		}
		else {
			dp[i][1]=max({dp[i-2][1],dp[i-2][0],dp[i-1][0]})+arr[i];
			dp[i][0]=arr[i-1]+dp[i-2][0];
		}
	}
	cout<<max(dp[n][0],dp[n][1])<<endl;
	return 0;
}