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 int N = 2e5+100;
ll f[N],g[N],h[N]; int a[N],n;

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    for(int i=2;i<=n;i+=2){
        f[i]=f[i-2]+a[i-1];
        g[i]=max(f[i-2],g[i-2])+a[i];
    }
    if(n&1){
        for(int i=3;i<=n;i+=2){
            h[i]=max(h[i-2]+a[i],max(f[i-1],g[i-1]));
        }printf("%lld\n",h[n]);
    }else printf("%lld\n",max(f[n],g[n]));
}