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 {
    static Map<Long, Long> mp = new HashMap<>();
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long[] A = new long[N];
        for(int i=0;i<N;i++){
            A[i] = sc.nextLong();
        }
        System.out.println(helper(A, 0, N/2));
    }
    private static long helper(long[] A, int ofs, int cnt){
        long key = ofs * 3_000_00 + cnt;
        if(mp.containsKey(key)){
            return mp.get(key);
        }
        if(cnt == 0){
            return 0;
        }
        int size = (A.length-ofs);
        if(size == 1 && cnt == 1){
            return A[ofs];
        }
        if(size == 2 && cnt == 1){
            return Math.max(A[ofs], A[ofs+1]);
        }
        if(size % 2 == 1 && size/2+1 == cnt){
            long ans = A[ofs] + helper(A, ofs+2, cnt-1);;
            mp.put(key, ans);
            return ans;
        }
        long use = A[ofs] + helper(A, ofs+2, cnt-1);
        long nouse = helper(A, ofs+1, cnt);
        long ans = Math.max(use, nouse);;
        mp.put(key,ans);
        return ans;
    }
}
