#include <bits/stdc++.h>
using namespace std;
int N, A;
queue<long long> que;
int main () {
    que.push(0);
    que.push(0);
    scanf("%d%d", &N, &A);
    long long x = A;
    for (int i = 2; i <= N; i++) {
        scanf("%d", &A);
        if (i % 2 == 0) {
            que.push(max(que.front()+A, x));
        }
        else {
            que.push(max(que.front()+A, que.back()));
            x += A;
        }
        que.pop();
    }
    printf("%lld\n", que.back());
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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, A;
queue<long long> que;
int main () {
    que.push(0);
    que.push(0);
    scanf("%d%d", &N, &A);
    long long x = A;
    for (int i = 2; i <= N; i++) {
        scanf("%d", &A);
        if (i % 2 == 0) {
            que.push(max(que.front()+A, x));
        }
        else {
            que.push(max(que.front()+A, que.back()));
            x += A;
        }
        que.pop();
    }
    printf("%lld\n", que.back());
}
```