#include <iostream>
#include <vector>

using namespace std;

const long long inf = 1000000000000000000;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;

	vector<long long> a;
	a.resize(n);

	for(int i = 0; i < n; ++i) {
		cin >> a[i];
	}

	vector<long long> dp;
	dp.resize(n);

	vector<long long> sum;
	sum.resize(n);

	for(int i = 0; i < n; ++i) {
		sum[i] = (i >= 2 ? sum[i - 2] : 0) + a[i];
	}
	
	dp[1] = max(a[0], a[1]);

	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 - 1] << '\n';

	return 0;
}

### Prompt

Develop a solution in Cpp 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 <vector>

using namespace std;

const long long inf = 1000000000000000000;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;

	vector<long long> a;
	a.resize(n);

	for(int i = 0; i < n; ++i) {
		cin >> a[i];
	}

	vector<long long> dp;
	dp.resize(n);

	vector<long long> sum;
	sum.resize(n);

	for(int i = 0; i < n; ++i) {
		sum[i] = (i >= 2 ? sum[i - 2] : 0) + a[i];
	}
	
	dp[1] = max(a[0], a[1]);

	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 - 1] << '\n';

	return 0;
}

```