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
n=int(input())
a=list(map(int,input().split()))
if n<=3:print(max(a));exit()
if n&1:
  dp=[3*[-(10**20)]for _ in range(n)]
  dp[0][0]=a[0]
  for i in range(1,n):
    if i%2==0:
      dp[i][0]=dp[i-2][0]+a[i]
      if i>=3:dp[i][2]=max(dp[i-2][2]+a[i],dp[i-3][1]+a[i])
      else:dp[i][2]=a[i]
    else:
      if i>=3:dp[i][1]=max(dp[i-2][1]+a[i],dp[i-3][0]+a[i])
      else:dp[i][1]=a[i]
  print(max(dp[-1][2],dp[-2][1],dp[-3][0]))
else:
  b=[a[0]]
  c=[a[1]]
  for i in range(2,n):
    if i%2==0:b.append(b[-1]+a[i])
    else:c.append(c[-1]+a[i])
  ans=max(b[-1],c[-1])
  for i in range(n//2-1):
    ans=max(ans,b[i]+c[-1]-c[i])
  print(ans)
