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
from sys import stdin

input = stdin.readline

n = int(input())
a = [int(x) for x in input().split()]
dp = [[0] * 3 for x in range(n)]

dp[0][0] = a[0]
dp[1][1] = a[1]


if n > 2:
    dp[2][0] = a[0] + a[2]
    dp[2][2] = a[2]

if n > 3:
    dp[3][1] = max(a[0] + a[3], a[1] + a[3])

for x in range(4, n):
    if x % 2 == 0:
        dp[x][0] = dp[x - 2][0] + a[x]
    dp[x][1] = max(dp[x - 2][1] + a[x], dp[x - 3][0] + a[x])
    dp[x][2] = max(dp[x - 2][2] + a[x], dp[x - 3][1] + a[x], dp[x - 4][0] + a[x])


if n % 2 == 0:
    print(max(dp[-2][0], dp[-1][1]))
else:
    print(max(dp[-3][0], dp[-2][1], dp[-1][2]))