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 ll long long int
vector<ll> v;
map<pair<int,int>, ll> mp;
int n;
ll solve(int st, int en, int cnt){
    if(cnt==0){
        return 0;
    }
    if((en-st+1)<(2*cnt-1) || en<st){
        return -(1ll<<60);
    }
    if(mp.count({st,cnt})) return mp[{st,cnt}];
    mp[{st,cnt}] = max(solve(st+1,en,cnt), v[st]+solve(st+2,en,cnt-1));
    return mp[{st,cnt}];
}
int main()
{
    cin >> n;
    v.resize(n);
    for(int i=0;i<n;i++) {cin >> v[i];}
    ll f = solve(0,n-1,n/2);
    cout << f << endl;
}
