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>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = int64_t;

int main(){
    int n; cin >> n;
    int a[n]; rep(i, n) cin >> a[i];

    int k = 1+n%2;
    ll dp[n+1][4];
    rep(i, n+1)rep(j, k+1) dp[i][j] = INT64_MIN/2;
    dp[0][0] = 0;
    rep(i, n){
        rep(j, k+1){
            dp[i+1][j+1] = max<ll>(dp[i+1][j+1], dp[i][j]);
            ll now = dp[i][j];
            if((i+j)%2 == 0) now += a[i];
            dp[i+1][j] = max<ll>(dp[i+1][j], now);
        }
    }
    cout << dp[n][k] << endl;

    return 0;
}