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 ll long long
const int N=2e5+5,mod=1e9+7;
const ll INF=1e15;
bool vis[N][5];
ll dp[N][5];
int a[N];
int n;
ll solve(int idx,int f)
{
    if(idx>=n)
    {
        if(f==2) return -INF;
        return 0;
    }
    ll &ret=dp[idx][f];
    if(vis[idx][f]==1) return ret;
    vis[idx][f]=1;
    ret=solve(idx+2,f)+a[idx];
    if(f) ret=max(ret,solve(idx+1,f-1));
    return ret;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i];
    cout<<solve(0,n%2+1)<<endl;
    return 0;
}
