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+5;
ll a[N],prfx[N],dp[N];
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n;cin>>n;
    for(ll i=1;i<=n;i++)cin>>a[i];
    prfx[1]=a[1];
    for(ll i=3;i<=n;i+=2)prfx[i]=prfx[i-2]+a[i];
    for(ll i=2;i<=n;i++){
        if(i&1)dp[i]=max(dp[i-1],dp[i-2]+a[i]);
        else dp[i]=max(prfx[i-1],a[i]+dp[i-2]);
    }
    cout<<dp[n];
}
