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

const int N = 2e5 + 5;
typedef long long ll;

ll a[N], ans, sum[N], f[N];
int n;


int main() {
	cin>>n; 
	for(int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
		if(i > 1) sum[i] = sum[i - 2] + a[i];
		else sum[i] = a[i];
	}
	for(int i = 2; i <= n; i++) {
		if(i&1) f[i] = max(f[i - 2] + a[i], f[i - 1]);
		else f[i] = max(f[i - 2] + a[i], sum[i - 1]); 
 	}
 	cout<<f[n]<<endl;
}

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

const int N = 2e5 + 5;
typedef long long ll;

ll a[N], ans, sum[N], f[N];
int n;


int main() {
	cin>>n; 
	for(int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
		if(i > 1) sum[i] = sum[i - 2] + a[i];
		else sum[i] = a[i];
	}
	for(int i = 2; i <= n; i++) {
		if(i&1) f[i] = max(f[i - 2] + a[i], f[i - 1]);
		else f[i] = max(f[i - 2] + a[i], sum[i - 1]); 
 	}
 	cout<<f[n]<<endl;
}

```