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
# dpでいけそう
n = int(input())
a = list(map(int, input().split()))
INF = 10**16

ans = -INF

a = [0, 0, 0, 0] + a
dp = [[-INF for _ in range(n+4)] for _ in range(3)]
dp[0][2] = 0
for i in range(4, len(a)):
    dp[2][i] = max(dp[2][i], dp[1][i-3] + a[i], dp[0][i-4] + a[i], dp[2][i-2] + a[i])
    dp[1][i] = max(dp[1][i], dp[0][i-3] + a[i], dp[1][i-2] + a[i])
    dp[0][i] = max(dp[0][i], dp[0][i-2] + a[i])
if n % 2 == 1:
    ans = max(ans, dp[0][n])
    ans = max(ans, dp[1][n+2])
    ans = max(ans, dp[2][n+3])
else:
    ans = max(ans, dp[0][n+2])
    ans = max(ans, dp[1][n+3])
# print(dp)
print(ans)