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>

const int N = 2e5 + 10;
typedef long long ll;

ll sum[N], dp[N], a[N];
int n, m, i, j, k;

int main(){
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    if (i & 1) {
      sum[i] = sum[i - 2] + a[i];
    }
  }
  dp[1] = 0;
  for (int i = 2; i <= n; i++) {
    if (i & 1) {
      dp[i] = std::max(dp[i - 2] + a[i], dp[i - 1]);
    } else {
      dp[i] = std::max(sum[i - 1], dp[i - 2] + a[i]);
    }
  }
  printf("%lld\n", dp[n]);
  return 0;  
}