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
#define pb push_back
#define N 200009
#define inf 1e18
int mod=998244353;


int a[N];
int n;
map<pair<int,int>,int>dp;
int fun(int idx,int pick){
    if(pick>n/2 or pick+(n-idx+1)/2 <n/2)return -inf;
    if(idx>=n)return 0;
    if(dp.count({idx,pick}))return dp[{idx,pick}];
    int x=fun(idx+1,pick);
    int y=fun(idx+2,pick+1)+a[idx];
    dp[{idx,pick}]=max(x,y);
    return max(x,y);
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin>>n;
    for(int i=0;i<n;i++)cin>>a[i];
    cout<<fun(0,0)<<"\n";
    return 0;
}