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
# coding: utf-8

N=int(input())
A=list(map(int,input().split()))
INF=10**18

dp=[[-INF for i in range(2)] for j in range(N+1)]
dp[0][0]=0
dp[1][0]=0
dp[2][0]=0
dp[1][1]=A[0]
dp[2][1]=A[1]

for i in range(3,N+1):
    dp[i][0]=max(dp[i-2][0]+A[i-1],dp[i-2][1])
    if i%2==0:
        dp[i][0]=max(dp[i][0],dp[i-3][0]+A[i-1])
    else:
        dp[i][0]=max(dp[i][0],dp[i-3][1]+A[i-1])
    
    dp[i][1]=dp[i-2][1]+A[i-1]
    if i%2==0:
        dp[i][1]=max(dp[i][1],dp[i-3][1]+A[i-1])

if N%2==0:
    print(max(dp[N][1],dp[N-1][1]))
else:
    print(max(dp[N][0],dp[N-1][1],dp[N-2][1]))
