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>
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define rep1(i,n) for(int i=1;i<=(int)n;i++)
#define sp(n) cout << fixed << setprecision(n)
typedef long long ll;
using namespace std;
int main(void){
    int n;cin>>n;
    vector<ll> a(n+1);
    rep1(i,n) cin>>a[i];
    ll sum[n+1],dp[n+1];
    rep(i,n+1){
        sum[i]=0;
        dp[i]=0;
    }
    
    sum[1]=a[1];
    for(int i=3;i<=n;i+=2) sum[i]=sum[i-2]+a[i];
    for(int i=2;i<=n;i++){
        if(i%2==1) dp[i]=max(dp[i-1],dp[i-2]+a[i]);
        else dp[i]=max(sum[i-1],dp[i-2]+a[i]);
        //cout<<dp[i]<<endl;
    }
    cout<<dp[n]<<endl;
}
