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 ll long long


int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	ll n;cin>>n;
	ll i,j;
	ll v[n];
	vector<vector<ll>> dp(n,vector<ll>(2,0));
	for(i=0;i<n;i++) cin>>v[i];
		//dp[0]={v[0],v[0],1};
		dp[0][0]=0;
		dp[0][1]=0;
		dp[1][0]=v[0];
	dp[1][1]=v[1];

		for(i=2;i<n;i++)
		{	if(i%2==0){
			dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
			dp[i][1]=max(dp[i-2][0],dp[i-2][1])+v[i];
		}
			else {dp[i][0]=dp[i-2][0]+v[i-1];
				dp[i][1]=(dp[i-1][0])+v[i];
			}

		}
		cout<<max(dp[n-1][0],dp[n-1][1]);
	return 0;
} 