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 rep2(i, m, n) for(int i=int(m); i<int(n); ++i)
#define drep2(i, m, n) for(int i=int(m-1); i>=int(n); --i)
#define rep(i, n) rep2(i, 0, n)
#define drep(i, n) drep2(i, n, 0)
using ll = long long;
using Vll = vector<ll>;
template<typename T> istream &operator>>(istream &is, vector<T> &v) { for (auto &e : v) is >> e; return is; }


int main() {
  ll n; cin >> n;
  Vll a(n); cin >> a;

  ll m = n/2;
  ll k = (n%2) ? 3 : 2;

  Vll dp(k);
  rep(i, m) drep(j, k) dp[j] = *max_element(dp.begin(), dp.begin()+j+1) + a[2*i+j];

  cout << *max_element(dp.begin(), dp.end()) << endl;
  return 0;
}
