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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<long long> A(N);
    for (long long &a: A)
        cin>>a;

    int d = N%2==0 ? 2 : 3;
    vector<vector<long long>> T(d, vector<long long>(N/2, -1000000000000000LL));
    for (int i=0; i<d; i++)
        T[i][0] = A[i];
    for (int i=1; i<N/2; i++)
        for (int j=0; j<d; j++)
            for (int k=0; k<=j; k++)
                T[j][i] = max(T[j][i], T[k][i-1]+A[2*i+j]);
    long long ans = -1000000000000000LL;
    for (int i=0; i<d; i++)
        ans = max(ans, T[i][N/2-1]);
    cout<<ans<<endl;
}
