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 int   long long
#define rep(i,a,b)  for(int i=a;i<b;i++)
#define m 1000000000000000007LL
int n,a[200005],dp[200005][4][2];
bool vis[200005][4][2];
int f(int pos,int cnt,int last){
	if(cnt<-1) return -m;
	if(pos==n) return 0;
	if(vis[pos][cnt+2][last]) return dp[pos][cnt+2][last];
	vis[pos][cnt+2][last]=1;
	int &ans=dp[pos][cnt+2][last];
	ans=f(pos+1,cnt-1,0);
	if(last==0) ans=max(ans,f(pos+1,cnt+1,1)+a[pos]);
	return ans;
}
void solve(){
    cin>>n;
    rep(i,0,n) cin>>a[i];
    cout<<f(0,n%2,0)<<endl;
}
 
signed main(){
	solve();
	
	
	
}