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 <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
int N;
vector<int> A;
unordered_map<int, int64_t> memo[3];
int64_t dfs(int idx, int skip) {
  if(idx >= N) return 0;
  if (memo[skip].count(idx)) return memo[skip][idx];

  int64_t ret = dfs(idx + 2, skip) + A[idx];
  if(skip > 0) ret = max(ret, dfs(idx + 1, skip-1));
  memo[skip][idx] = ret;
  return ret;
}

int main(void) {
  ios::sync_with_stdio(false);
  cin >> N;
  A.resize(N);
  for(int i = 0; i < N; ++i) cin >> A[i];
  int skip = (N % 2 == 0) ? 1 : 2;
  cout << dfs(0, skip) << endl;  
  return 0;
}
