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;
const int Max=200000;
long A[Max]={}, f[Max]={}, seen[Max]={}, E[Max]={0};
long F(int n){
long M=0;
if(n==2){ M=max(A[0], A[1]); return M; }
if(n==3){ M=max(F(2), A[2]); return M; }
if(seen[n]){ return f[n]; }
seen[n]=1; 
if(n%2){ return f[n]=max(F(n-1), F(n-2)+A[n-1]); }
else{ return f[n]=max(E[n-2], F(n-2)+A[n-1]); }
}

int main(){
int N, i; cin >> N;
for(i=0; i<N; i++){ cin >> A[i]; }
E[0]=A[0];
for(i=2; i<N; i+=2){ E[i]=E[i-2]+A[i]; }  
cout << F(N) << "\n";
}