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;

int main(){
    //freopen("a.in", "r", stdin);
    int n; cin >> n;
    int arr[n];
    for(int i = 0; i<n; i++) cin >> arr[i];
    long long dp[n];
    dp[0] = 0;
    dp[1] = max(arr[0], arr[1]);
    long long pre[n];
    pre[0] = arr[0];
    for(int i = 2; i<n; i+=2) pre[i] = pre[i-2]+arr[i];
    for(int i = 2; i<n; i++){
        if(i % 2 == 0){
            dp[i] = max(arr[i]+dp[i-2], dp[i-1]);
        }
        else{
            dp[i] = max(arr[i]+dp[i-2], pre[i-1]);
        }
    }
    cout << dp[n-1] << endl;
}