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 FOR(i,N) for(int i=0; i < (int)N; i++)
int N;
ll A[202020];
 
ll dp[202020][2][3];
 
void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	
	FOR(i,N+1) FOR(x,2) FOR(y,3) dp[i][x][y]=-1LL<<60;
	dp[0][1][0]=0;
	
	FOR(i,N) {
		cin>>x;
		FOR(j,2) FOR(k,3) {
			// not take
			if(k+j<3) dp[i+1][1][k+j]=max(dp[i+1][1][k+j],dp[i][j][k]);
			// take
			if(j==1) dp[i+1][0][k]=max(dp[i+1][0][k],dp[i][j][k]+x);
		}
	}
	
	if(N%2==0) cout<<max(dp[N][1][0],dp[N][0][1])<<endl;
	else cout<<max(dp[N][0][2],dp[N][1][1])<<endl;
}

int main(){
	solve();
}