#include <iostream>
#include <cstring>
using namespace std;
long long a[200005], sum[200005];
long long dp[200005];
int main() {
    int n;
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i], sum[i] = a[i] + (i > 1? sum[i - 2]: 0);
	for (int i = 1; i < n; i++) {
		if (i % 2 == 0)
			dp[i] = max(a[i] + dp[i - 2], dp[i - 1]);
		else
			dp[i] = max(a[i] + dp[i - 2], sum[i - 1]);
	}
    cout << dp[n - 1] << endl;
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <iostream>
#include <cstring>
using namespace std;
long long a[200005], sum[200005];
long long dp[200005];
int main() {
    int n;
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i], sum[i] = a[i] + (i > 1? sum[i - 2]: 0);
	for (int i = 1; i < n; i++) {
		if (i % 2 == 0)
			dp[i] = max(a[i] + dp[i - 2], dp[i - 1]);
		else
			dp[i] = max(a[i] + dp[i - 2], sum[i - 1]);
	}
    cout << dp[n - 1] << endl;
	return 0;
}
```