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

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	int n;
	cin >> n;
	vector<long long> a(n + 1), dp(n + 1, 0), pre(n + 1, 0);
	for (int i = 1; i <= n; ++i){
		cin >> a[i];
	}
	pre[1] = a[1];
	for (int i = 2; i <= n; ++i){
		pre[i] = pre[i - 2] + a[i];
	}
	dp[1] = 0;
	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(dp[i - 2] + a[i], pre[i - 1] );
		}
	}
	cout << dp[n];
	return 0;
}


### Prompt

Create a solution in Cpp for 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 main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	int n;
	cin >> n;
	vector<long long> a(n + 1), dp(n + 1, 0), pre(n + 1, 0);
	for (int i = 1; i <= n; ++i){
		cin >> a[i];
	}
	pre[1] = a[1];
	for (int i = 2; i <= n; ++i){
		pre[i] = pre[i - 2] + a[i];
	}
	dp[1] = 0;
	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(dp[i - 2] + a[i], pre[i - 1] );
		}
	}
	cout << dp[n];
	return 0;
}


```