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;
#define fi first
#define se second
#define pb push_back
#define forn(i,n) for(int i=0;i<(n);i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forr(i,n) for(int i=n;i>=0;i--)
#define all(x) x.begin(), x.end()
const int MAXN = 2e5 +5;

void fio(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}

LL dp[MAXN], odds[MAXN];

int main(){
  fio();
  int n; cin>>n;
  vector<LL> v(n);
  forn(i,n) cin>>v[i];
  forn(i,n) odds[i] = ( i%2 || i<2 ?0:odds[i-2]) + v[i];
  for(int i=1;i<n;i++)
  	dp[i] = max((i>=2?dp[i-2]:0)+v[i], i%2?odds[i-1]:dp[i-1]);
  cout<<dp[n-1]<<endl;
  return 0;
}