#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 5;

int n;
ll a[MAXN],dp[MAXN];

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	scanf("%d",&n);
	for(int i = 1;i <= n;i++)
		scanf("%lld",&a[i]);
	ll sum = a[1];
	for(int i = 2;i <= n;i++){
		if(i & 1){
			sum += a[i];
			dp[i] = max(dp[i - 1],dp[i - 2] + a[i]);
		}else
			dp[i] = max(sum,dp[i - 2] + a[i]);
	}
	printf("%lld\n",dp[n]);
	return 0;
}


### Prompt

Develop a solution in CPP 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>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 5;

int n;
ll a[MAXN],dp[MAXN];

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	scanf("%d",&n);
	for(int i = 1;i <= n;i++)
		scanf("%lld",&a[i]);
	ll sum = a[1];
	for(int i = 2;i <= n;i++){
		if(i & 1){
			sum += a[i];
			dp[i] = max(dp[i - 1],dp[i - 2] + a[i]);
		}else
			dp[i] = max(sum,dp[i - 2] + a[i]);
	}
	printf("%lld\n",dp[n]);
	return 0;
}


```