#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
const int mxn = 2e5 + 1;
ll arr[mxn], dp[mxn], pref[mxn];
 
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;cin>>n;
	for(int i = 1; i<=n; ++i) cin>>arr[i];
	dp[0] = 0;
	pref[0] = 0;
	pref[1] = arr[1];
	for(int i = 2; i <=n; ++i) pref[i] = pref[i-2] + arr[i];
	dp[1] = 0;
	for(int i = 2; i <=n; ++i){
		if(i&1){
			dp[i] = max(dp[i-1], dp[i-2] + arr[i]);
		} else{
			dp[i] = max(dp[i-2] + arr[i], pref[i-1]);
		}
	}
	cout<<dp[n]<<endl;
	return 0;
}
 
### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
const int mxn = 2e5 + 1;
ll arr[mxn], dp[mxn], pref[mxn];
 
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;cin>>n;
	for(int i = 1; i<=n; ++i) cin>>arr[i];
	dp[0] = 0;
	pref[0] = 0;
	pref[1] = arr[1];
	for(int i = 2; i <=n; ++i) pref[i] = pref[i-2] + arr[i];
	dp[1] = 0;
	for(int i = 2; i <=n; ++i){
		if(i&1){
			dp[i] = max(dp[i-1], dp[i-2] + arr[i]);
		} else{
			dp[i] = max(dp[i-2] + arr[i], pref[i-1]);
		}
	}
	cout<<dp[n]<<endl;
	return 0;
}
 
```