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;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e18;

vec l(2e5+50),r(2e5+50);
vec a(2e5+50);

ll cal(ll x) {
  if(x<3) {
    return 0;
  }
  return max(max(l[x-1],r[x-1]),cal(x-2)+a[x-1]);
}

int main() {
  ll n;
  cin >> n;
  for(ll i=0;i<n;i++) cin >> a[i];
  for(ll i=2;i<=n;i+=2){
    l[i]=l[i-2]+a[i-2];
    r[i]=max(r[i-2],l[i-2])+a[i-1];
  }
  if(n%2) {
    cout << cal(n) << endl;
  }
  else {
    cout << max(l[n],r[n]) << endl;
  }
}