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.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		String[] sa = br.readLine().split(" ");
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(sa[i]);
		}
		br.close();

		if (n == 2) {
			System.out.println(Math.max(a[0], a[1]));
			return;
		}

		long ans = Long.MIN_VALUE;
		long[][] dp = new long[n][3];
		dp[0][0] = a[0];
		dp[1][0] = a[0];
		dp[2][0] = a[0] + a[2];
		dp[1][1] = a[1];
		dp[2][1] = a[1];
		dp[2][2] = a[2];
		for (int i = 3; i < n; i++) {
			if (i % 2 == 0) {
				dp[i][0] = dp[i - 2][0] + a[i];
				dp[i][1] = dp[i - 1][1];
				dp[i][2] = Math.max(dp[i - 2][2], dp[i - 3][1]) + a[i];
			} else {
				dp[i][0] = dp[i - 1][0];
				dp[i][1] = Math.max(dp[i - 2][1], dp[i - 3][0]) + a[i];
				dp[i][2] = dp[i - 1][2];
			}
		}
//		for (int i = 0; i < dp.length; i++) {
//			for (int j = 0; j < dp[i].length; j++) {
//				System.out.print(dp[i][j] + "\t");
//			}
//			System.out.println();
//		}
		if (n % 2 == 0) {
			ans = Math.max(ans, dp[n - 2][0]);
			ans = Math.max(ans, dp[n - 1][1]);
		} else {
			ans = Math.max(ans, dp[n - 3][0]);
			ans = Math.max(ans, dp[n - 2][1]);
			ans = Math.max(ans, dp[n - 1][2]);
		}
		System.out.println(ans);
	}
}
