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()))
m=-10**15
#dp[i][j]=i番目までで、j回選ばない連続を使ったときの最大値
dp1=[[m,m,m] for i in range(n)] #最後を使うとき
dp0=[[m,m,m] for i in range(n)] #最後を使わないとき
dp1[0][0]=A[0]
dp0[0][1]=0 #最初選ばないときはj=1にする
for i in range(1,n):
    for j in range(3):
        dp1[i][j]=max(dp1[i][j],dp0[i-1][j]+A[i])
        dp0[i][j]=max(dp0[i][j],dp1[i-1][j])
        if j!=2:
            dp0[i][j+1]=max(dp0[i][j+1],dp0[i-1][j])
if n%2==0:
    print(max(dp1[n-1][1],dp0[n-1][0]))
else:
    print(max(dp1[n-1][2],dp0[n-1][1]))