#include<bits/stdc++.h>
using namespace std;
using ll=long long;
ll dat[1<<18];
ll sum[1<<18];
ll dp[1<<18];
ll n;
int main(){
	cin>>n;
	for(ll i=0;i<n;i++) {
		cin>>dat[i];
		if(i>=2 and i%2==0) sum[i]=dat[i]+sum[i-2];
		else if(i==0) sum[i]=dat[i];
	}
	for(ll i=0;i<n;i++){
		if(i&1){
			dp[i]=max(dp[i-2]+dat[i] , sum[i-1]);
		}
		else{
			dp[i]=max(dp[i-2]+dat[i] , dp[i-1]);
		}
	}
	cout<<dp[n-1]<<endl;
	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<bits/stdc++.h>
using namespace std;
using ll=long long;
ll dat[1<<18];
ll sum[1<<18];
ll dp[1<<18];
ll n;
int main(){
	cin>>n;
	for(ll i=0;i<n;i++) {
		cin>>dat[i];
		if(i>=2 and i%2==0) sum[i]=dat[i]+sum[i-2];
		else if(i==0) sum[i]=dat[i];
	}
	for(ll i=0;i<n;i++){
		if(i&1){
			dp[i]=max(dp[i-2]+dat[i] , sum[i-1]);
		}
		else{
			dp[i]=max(dp[i-2]+dat[i] , dp[i-1]);
		}
	}
	cout<<dp[n-1]<<endl;
	return 0;
}

```