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>
#define DEBUG if(0)
#define lli long long int
#define ldouble long double
using namespace std;

const int maxN = 2e5; int n;
lli a[maxN];

lli memo[maxN][3];
lli solve(int i = 0, int jumps = n & 1 ? 2 : 1)
{
  if (i >= n) return 0;
  if (memo[i][jumps] != -1) return memo[i][jumps];

  lli ans = a[i] + solve(i + 2, jumps);
  if (jumps) ans = max(ans, solve(i + 1, jumps - 1));
  return memo[i][jumps] = ans;
}

int main()
{
  while (scanf("%d", &n) != EOF)
  {
    for (int i = 0; i < n; i++)
      scanf("%lld", &a[i]);

    memset(memo, -1, sizeof(memo));
    lli ans = solve();
    printf("%lld\n", ans);
  }
  return 0;
}