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 <stdio.h>
#define mod 1000000007
typedef long long ll;

int main(void) {
  ll i, j, k, n, ans = 0;
  scanf("%lld", &n);
  ll a[n], dp[n][2];
  for(i = 0; i < n; ++i) scanf("%lld", &a[i]), dp[i][0] = dp[i][1] = -1e18;
  dp[0][1] = 0, dp[1][1] = 0;
  dp[0][0] = a[0];
  dp[1][0] = a[0] > a[1] ? a[0] : a[1];
  for(i = 2; i < n; ++i) {
    if(i & 1) dp[i][0] = dp[i - 1][0], dp[i][1] = dp[i - 1][1];
    else dp[i][1] = dp[i - 1][0];
    if(dp[i][0] < dp[i - 2][0] + a[i]) dp[i][0] = dp[i - 2][0] + a[i];
    if(dp[i][1] < dp[i - 2][1] + a[i]) dp[i][1] = dp[i - 2][1] + a[i];
  }
  printf("%lld", dp[n - 1][n & 1]);
  return 0;
}