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
#include <iostream>
using namespace std;
#define el '\n'
typedef long long ll;
const int MAXN = 2e5+2;
int N, A[MAXN];
ll odd[MAXN], dp[MAXN];
int main(){
	cin.tie(0);
	cin.sync_with_stdio(0);
	cin >> N;
	for(int i = 1;i<=N;i++) cin >> A[i];
	odd[1] = A[1];
	for(int i = 2;i<=N;i++){
		if(i%2){
			odd[i] = odd[i-2]+A[i];
		}
	}
	dp[1] = 0;
	for(int i = 2;i<=N;i++){
		if(i%2){
			dp[i] = max(dp[i-2]+A[i], dp[i-1]);
		}else{
			dp[i] = max(dp[i-2]+A[i], odd[i-1]);
		}
	}
	cout << dp[N] << el;
}