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;


using ll = long long;
const int MaxN = 2e5+10;
int N;
int A[MaxN];


void solve()
{
    vector<ll> prefix(N + 1);
    vector<ll> f(N + 1);
    prefix[1] = A[1];
    for (int i = 3; i <= N; i += 2)
        prefix[i] = prefix[i - 2] + A[i];
    for (int n = 2; n <= N; n++) {
        if (n & 1)
            f[n] = max(f[n - 1], A[n] + f[n - 2]);
        else
            f[n] = max(prefix[n - 1], A[n] + f[n - 2]);
    }
    cout << f[N] << endl;
}


int main()
{
    cin >> N;
    for (int i = 1; i <= N; i++) cin >> A[i];
    solve();
    return 0;
}
