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.Scanner;

public class Main {

	public static void main(String[] args) { //No MLE!
		Scanner input = new Scanner(System.in);
		int N = input.nextInt(); 
		long[][] dp;
		long[] arr = new long[N];
		for (int i = 0; i < N; i++) {
			arr[i] = input.nextLong();
		}
		if (N%2==0) {
			dp = new long[N/2][2];
			for (int i = 0; i < N/2; i++) {
				for (int j = 0; j < 2; j++) {
					if (i==0&&j==0) {
						dp[i][j] = arr[i*2+j];
					}
					else if (j==0) {
						dp[i][j] = dp[i-1][j]+arr[i*2];
					}
					else if (i>0){
						dp[i][j] = dp[i][j-1];
						long curNum = arr[i*2+1];
						dp[i][j] = Math.max(dp[i][j], curNum+Math.max(dp[i-1][j-1],dp[i-1][j]));
					}else if (i==0) {
						dp[i][j] = Math.max(dp[i][j-1],arr[i*2+1]);
					}	
				}
			}
			/*for (int i = 0; i < N/2; i++) {
				for (int j = 0; j < 2; j++) {
					System.out.print(dp[i][j]+" ");
				}
				System.out.print("\n");
			}*/
			System.out.println(dp[N/2-1][1]);
		}
		else {
			dp = new long[N/2][3];
			for (int i = 0; i < N/2; i++) {
				for (int j = 0; j < 3; j++) {
					if (j==0&&i==0) dp[i][j] = dp[i][j] = arr[0];
					else if (j==0) {
						dp[i][j] = dp[i-1][j]+arr[i*2];
					}
					else if (i==0){
						dp[i][j] = Math.max(dp[i][j-1],arr[i*2+j]);
					}else if (i>0) {
						dp[i][j] = dp[i][j-1];
						long curNum = arr[i*2+j];
						dp[i][j] = Math.max(dp[i][j], dp[i-1][j]+curNum);
					}
				}
			}
			/*for (int i = 0; i < N/2; i++) {
				for (int j = 0; j < 3; j++) {
					System.out.print(dp[i][j]+" ");
				}
				System.out.print("\n");
			}*/
			System.out.println(dp[N/2-1][2]);
		}
	}
}