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
INF = 10**18

def solve(n, a):
    # 現在の位置 x 選んだ個数 x 直前を選んだかどうか
    dp = [{j: [-INF, -INF] for j in range(i//2-1, (i+1)//2 + 1)} for i in range(n+1)]
    dp[0][0][False] = 0
    for i in range(n):
        for j in dp[i].keys():
            if (j+1) in dp[i+1]:
                dp[i+1][j+1][True] = max(dp[i+1][j+1][True], dp[i][j][False] + a[i])
            if j in dp[i+1]:
                dp[i+1][j][False] = max(dp[i+1][j][False], dp[i][j][False])
                dp[i+1][j][False] = max(dp[i+1][j][False], dp[i][j][True])
    return max(dp[n][n//2])

n = int(input())
a = list(map(int, input().split()))
print(solve(n, a))