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 sys
input=lambda: sys.stdin.readline().rstrip()
n=int(input())
A=[int(i) for i in input().split()]
inf=float("inf")
DP=[[-inf]*4 for _ in range(n+1)]
DP[0][2]=0
for i,a in enumerate(A):
  if (i+1)%2==0:
    DP[i+1][0]=DP[i][3]+a
    if i>=1:
      DP[i+1][0]=max(DP[i+1][0],DP[i-1][2]+a)
    DP[i+1][2]=DP[i][0]
    DP[i+1][3]=max(DP[i][1],DP[i][3])
    if i>=1:
      DP[i+1][3]=max(DP[i+1][3],DP[i-1][2])
  else:
    DP[i+1][0]=DP[i][2]+a
    DP[i+1][1]=DP[i][3]+a
    DP[i+1][3]=max(DP[i][0],DP[i][2])
if n%2==0:
  print(max(DP[n][0],DP[n][2]))
else:
  print(max(DP[n][1],DP[n][3]))

