#include <bits/stdc++.h>
using namespace std;
int N;
long long A[200001], dp[200001];
int main () {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
    dp[0] = dp[1] = 0;
    long long x = A[1];
    for (int i = 2; i <= N; i++) {
        if (i % 2 == 0) {
            dp[i] = max(dp[i-2]+A[i], x);
        }
        else {
            dp[i] = max(dp[i-2]+A[i], dp[i-1]);
            x += A[i];
        }
    }
    printf("%lld\n", dp[N]);
}
### Prompt

Your challenge is to write a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int N;
long long A[200001], dp[200001];
int main () {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
    dp[0] = dp[1] = 0;
    long long x = A[1];
    for (int i = 2; i <= N; i++) {
        if (i % 2 == 0) {
            dp[i] = max(dp[i-2]+A[i], x);
        }
        else {
            dp[i] = max(dp[i-2]+A[i], dp[i-1]);
            x += A[i];
        }
    }
    printf("%lld\n", dp[N]);
}
```