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 main(){

    int n;
    cin>>n;
    vector<int> arr(n+1);

    for(int i = 1; i <= n; i++) cin>>arr[i];

    vector<ll> pf(n+1);
    vector<ll> dp(n+1,LLONG_MIN);
    for(int i = 1; i <= n; i++) if(i&1) pf[i]+=arr[i]+pf[i-2 < 0 ? 0 : i-2]; // pref of odd sum

    dp[0] = 0;
    dp[1] = 0;
    for(int i = 2; i <= n; i++){
        if(i&1) dp[i] = max(dp[i-1],arr[i]+dp[i-2]); // for odd 
        else dp[i] = max(pf[i-1],dp[i-2]+arr[i]); //for even
    }
    cout<<dp[n];
}
