#include <bits/stdc++.h>
#define ll long long
#define intm INT_MAX
#define llm LLONG_MAX

using namespace std;
int MOD = 998244353;

ll ar[200009];
int N;
ll dp[200009];
ll odd[200009];

int main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> N;
	for(int i=1; i <= N; i++) cin >> ar[i];
	odd[1] = ar[1];
	for(int i=2; i <=N; i++){
        if(i%2!=0) odd[i] = odd[i-2]+ ar[i];
	}
	dp[0] = dp[1] = 0;
	for(int i=2; i <= N; i++){
        dp[i] = dp[i-2] + ar[i];
        if( i % 2 == 0 ) dp[i] =  max( dp[i], odd[i-1] );
        else{
            dp[i] = max(dp[i], dp[i-1] );
        }
	}
	cout << dp[N];

}

### 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>
#define ll long long
#define intm INT_MAX
#define llm LLONG_MAX

using namespace std;
int MOD = 998244353;

ll ar[200009];
int N;
ll dp[200009];
ll odd[200009];

int main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> N;
	for(int i=1; i <= N; i++) cin >> ar[i];
	odd[1] = ar[1];
	for(int i=2; i <=N; i++){
        if(i%2!=0) odd[i] = odd[i-2]+ ar[i];
	}
	dp[0] = dp[1] = 0;
	for(int i=2; i <= N; i++){
        dp[i] = dp[i-2] + ar[i];
        if( i % 2 == 0 ) dp[i] =  max( dp[i], odd[i-1] );
        else{
            dp[i] = max(dp[i], dp[i-1] );
        }
	}
	cout << dp[N];

}

```