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>
#define endl "\n"
using namespace std;
typedef long long int lli;

void solve(){
	int n;
	cin>>n;
	vector<lli> a(n+1);
	for(int i=1;i<n+1;i++){
		cin>>a[i];
	}
	vector<lli> dp(n+1,0);
	dp[1] = 0;
	vector<lli> prefix(n+1);
	prefix[1] = a[1];
	for(int i=3;i<=n;i+=2){
		prefix[i] = prefix[i-2]+a[i];
	}
	for(int i=2;i<=n;i++){
		if(i&1){
			dp[i]=max(dp[i-1],dp[i-2]+a[i]);
		}else{
			dp[i] = a[i] + dp[i-2];
			dp[i] = max(dp[i],prefix[i-1]);
		}
	}
	cout<<dp[n];
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    solve();//checking
    return 0;
}