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;
#define int long long

vector<map<int, int> >dp(1000000);
int arr[1000000];
int inf = 1e16;

int solve(int i, int k){
    if(i<0 && k==0)return 0;
    if(i<0)return -inf;
    int aa = (i+1+1)/2;
    if(k > aa)return -inf;
    if(dp[i].find(k) != dp[i].end())return dp[i][k];
    int alpha = max(solve(i-1, k), arr[i]+solve(i-2,k-1));
    return dp[i][k] = alpha;
}

int32_t main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n;
    cin>>n;
    for(int i=0; i<n; i++)cin>>arr[i];
    cout<<solve(n-1, n/2);
    return 0;
}