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
#include <bits/stdc++.h>
using namespace std;

const int N = 2e+5 + 5;
long long int a[N], dp[N][2];
int n;

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    dp[1][1] = a[1];
    for (int i = 2; i <= n; i++)
    {
        if (i % 2 == 0)
        {
            dp[i][1] = a[i] + max(dp[i - 2][0], dp[i - 2][1]);
            dp[i][0] = a[i - 1] + dp[max(0, i - 2)][0];
        }
        else
        {
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);
            dp[i][1] = a[i] + max(dp[i - 2][1], dp[i - 2][0]);
            if (i == 3)
                dp[i][1] = a[i];
        }
    }
    cout << max(dp[n][0], dp[n][1]);
}