#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int max_n = 200005;
int n, a[max_n];
ll dp[max_n];
ll sum[max_n];
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", a+i);
    sum[1] = a[1];
    for(int i = 2; i <= n; i++) {
        sum[i] = sum[i-2] + a[i]; 
    }
    for(int i = 2; i <= n; i++) {
        if(i&1) {
            dp[i] = max(dp[i-1], dp[i-2] + a[i]);
        }
        else {
            dp[i] = max(sum[i-1], a[i] + dp[i-2]);
        }
    }
    printf("%lld\n", dp[n]);
    return 0;
}
### 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;
typedef long long ll;
const int max_n = 200005;
int n, a[max_n];
ll dp[max_n];
ll sum[max_n];
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", a+i);
    sum[1] = a[1];
    for(int i = 2; i <= n; i++) {
        sum[i] = sum[i-2] + a[i]; 
    }
    for(int i = 2; i <= n; i++) {
        if(i&1) {
            dp[i] = max(dp[i-1], dp[i-2] + a[i]);
        }
        else {
            dp[i] = max(sum[i-1], a[i] + dp[i-2]);
        }
    }
    printf("%lld\n", dp[n]);
    return 0;
}
```