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 rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll dp[200001]={0};
int main(){
    ll n,A[200001],B[200002];
    cin>>n;
    rep(i,n) cin>>A[i];
    rep(i,n){
        if(i==1) B[2]=A[0];
        if(i%2&&i>1) B[i+1]=B[i-1]+A[i-1];
    }
    rep(i,n+1){
        if(i<2) continue;
        if(i%2==0) dp[i]=max(B[i],dp[i-2]+A[i-1]);
        else dp[i]=max(dp[i-2]+A[i-1],dp[i-1]);
    }
    //rep(i,n+1) cout<<dp[i]<<"\n";
    cout<<dp[n]<<"\n";
}