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;

typedef long long ll;

const ll inf=1e18;

int main(){
    int N; cin >> N;
    ll a[N];
    for(int i=0;i<N;i++){
        cin >> a[i];
    }
    int M=N%2+1;
    vector<vector<ll>> dp(N+10,vector<ll>(M+1,-inf));
    dp[0][0]=0;
    for(int i=0;i<=N;i++){
        for(int j=0;j<M;j++){
            dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]);
        }
        for(int j=0;j<=M;j++){
            dp[i+2][j]=max(dp[i+2][j],dp[i][j]+a[i]);
        }
    }
    cout << dp[N+1][M] << endl;
}