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;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
const ll MAX_N=2e5;
ll N,A[MAX_N],dp[MAX_N],psm[MAX_N];
int main(){
	cin.tie(0);
	cin.sync_with_stdio(false);
	cin>>N;
	REP(i,N){
		cin>>A[i];
	}
	psm[0]=A[0];
	for(ll i=2;i<N;i+=2){
		psm[i]=psm[i-2]+A[i];
	}
	dp[0]=0;
	dp[1]=max(A[0],A[1]);
	FOR(i,2,N){
		if(i%2==0){
			dp[i]=max(dp[i-1],A[i]+dp[i-2]);
		}else{
			dp[i]=max(psm[i-1],A[i]+dp[i-2]);
		}
	}
	cout<<dp[N-1]<<endl;
}