#include <iostream>
#include <cstdio>
using namespace std;
long long n;
long long a[200010];
long long dp[200010];
long long sum[200010];
int main() {
	scanf("%lld", &n);
	for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
	sum[1] = a[1];
	for (long long i = 3; i <= n; i += 2) {
		sum[i] = sum[i - 2] + a[i];
	}
	for (long long 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]);
		}
	}
	printf("%lld", dp[n]);
	return 0;
}
### Prompt

Your task is to create a cpp solution to 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 <iostream>
#include <cstdio>
using namespace std;
long long n;
long long a[200010];
long long dp[200010];
long long sum[200010];
int main() {
	scanf("%lld", &n);
	for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
	sum[1] = a[1];
	for (long long i = 3; i <= n; i += 2) {
		sum[i] = sum[i - 2] + a[i];
	}
	for (long long 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]);
		}
	}
	printf("%lld", dp[n]);
	return 0;
}
```