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<bits/stdc++.h>
#define st first
#define nd second

using ll = long long;
using namespace std;
using pii = pair<int, int>;
const int N = 2e5 + 10;
ll dp[N][2];
int a[N];
int main(){
	int n;
	scanf("%d", &n);
	for(int i = 1; i <=n; i++){
		scanf("%d", &a[i]);
	}
	dp[1][0] = 0LL;
	dp[1][1] = 1LL * a[1];
	dp[2][0] = 1LL * max(a[1], a[2]);
	for(int i = 3; i <= n; i++){
		if(i % 2 == 0){
			dp[i][0] = max(dp[i - 1][1], dp[i - 2][0] + 1LL * a[i]);
		}
		else{
			dp[i][0] = max(dp[i - 1][0], dp[i - 2][0] + 1LL * a[i]);
			dp[i][1] = dp[i - 2][1] + 1LL * a[i];
		}
	}
	printf("%lld", dp[n][0]);
}