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 int;

int main(void) {
    cin.tie(0); ios::sync_with_stdio(false);

    ll n;
    cin >> n;
    vector<ll> a(n);
    for(auto &e: a) cin >> e;
    vector<vector<ll>> dp(n, vector<ll>(50, -1e18));
    dp[0][1 + 25] = a[0];
    dp[0][-1 + 25] = 0;
    dp[1][0 + 25] = a[1];
    dp[1][-2 + 25] = 0;
    for(ll i=1; i<n; i++) {
        for(ll j=-3; j<=3; j++) {
            if(i-2>=0) dp[i][j + 25] = max(dp[i][j+1 + 25], dp[i-2][j + 25] + a[i]);
            dp[i][j-1 + 25] = max(dp[i][j-1 + 25], dp[i-1][j + 25]);
        }
    }
    cout << dp[n-1][n/2 - (n+1)/2 + 25] << endl;

}
