#include<bits/stdc++.h>
using namespace std;

#define ll long long
ll sum[200001], a[200001], dp[200001];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int n;
	cin >> n;
	for(int i=1;i<=n;i++) {
		cin >> a[i];
		sum[i] = (i > 2 ? sum[i-2] : 0) + 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], dp[i-2]+a[i]);
	}
	cout << dp[n];
}
### 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;

#define ll long long
ll sum[200001], a[200001], dp[200001];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int n;
	cin >> n;
	for(int i=1;i<=n;i++) {
		cin >> a[i];
		sum[i] = (i > 2 ? sum[i-2] : 0) + 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], dp[i-2]+a[i]);
	}
	cout << dp[n];
}
```