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
import java.util.*;

public class Main {
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		final long MIN = (long)-1e16;
		int N = sc.nextInt();
		int[] a = new int[N];
		for(int i=0; i<N; i++) {
			a[i] = sc.nextInt();
		}

		long[][][] dp = new long[N+1][3][2];
		for(int i=0; i<=N; i++)
			for(int j=0; j<3; j++)
				dp[i][j][0] = dp[i][j][1] = MIN;
		if(N%2==0) {
			dp[0][1][0] = 0;
		} else {
			dp[0][2][0] = 0;
		}
		
		for(int i=0; i<N; i++) {
			dp[i+1][0][0] = Math.max(dp[i+1][0][0], dp[i][1][0]);
			dp[i+1][1][0] = Math.max(dp[i+1][1][0], dp[i][2][0]);
			dp[i+1][0][0] = Math.max(dp[i+1][0][0], dp[i][0][1]);
			dp[i+1][1][0] = Math.max(dp[i+1][1][0], dp[i][1][1]);
			dp[i+1][2][0] = Math.max(dp[i+1][2][0], dp[i][2][1]);

			dp[i+1][0][1] = Math.max(dp[i+1][0][1], dp[i][0][0] + a[i]);
			dp[i+1][1][1] = Math.max(dp[i+1][1][1], dp[i][1][0] + a[i]);
			dp[i+1][2][1] = Math.max(dp[i+1][2][1], dp[i][2][0] + a[i]);
		}
		long ans = MIN;
		for(int i=0; i<3; i++) {
			for(int j=0; j<2; j++) {
				ans = Math.max(ans, dp[N][i][j]);
			}
		}
		System.out.println(ans);
		
		sc.close();
	}

}
