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>
#define pii pair<int,int>
#define pb push_back
#define nmax 200011
#define lld long long
#define inf 9999999999999999LL
using namespace std;
int a[nmax];
lld dp[nmax], st[nmax];
int main()
{
    int n;
    cin>>n;
    for (int i=1;i<=n;++i){
        cin>>a[i];
    }
    st[1] = a[1];
    for (int i=2;i<=n;++i){
        st[i] = st[i-2] + a[i];
    }
    for (int i=2;i<=n;++i){
        if (i % 2){
            dp[i] = max(dp[i-1], dp[i-2] + a[i]);
        } else {
            dp[i] = a[i] + dp[i-2];
            dp[i] = max(dp[i], st[i-1]);
        }
    }
    cout<<dp[n]<<'\n';
    return 0;
}
