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>
using namespace std;

const long INF = 1L << 60;
int a[200010];
long dp[200010][4];

template<class T> inline bool chmax(T &a, const T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

int main(){
	int n, a;
	scanf("%d", &n);
	int k = n % 2 + 1;
	for(int i = 0; i <= n; i++)
		for(int j = 0; j <= k; j++)
			dp[i][j] = -INF;
	dp[0][0] = 0;
	for(int i = 0; i < n; i++) {
		scanf("%d", &a);
		for(int j = 0; j <= k; j++) {
			chmax(dp[i+1][j+1], dp[i][j]);
          	if((i+j) % 2) 
				chmax(dp[i+1][j], dp[i][j]);
			else
				chmax(dp[i+1][j], dp[i][j] + a);
		}
	}
	printf("%ld\n", dp[n][k]);
}
