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 rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
typedef long long ll;

ll dp[200005][4];
ll INF = 1e+18;

int main(){
	ll n;
	cin >> n;
	vector<ll> a(n);
	rep(i,n) cin >> a[i];
	ll k = 1 + n % 2 ;
	rep(i,n+1)rep(j,k+1) dp[i][j] = -INF;
	dp[0][0] = 0;
	rep(i,n){
		rep(j,k+1){
			dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]);
			ll now = dp[i][j];
			if((i+j)%2 == 0) now += a[i];
			dp[i+1][j] = max(dp[i+1][j], now);
		}
	}
	ll ans = dp[n][k];
	cout << ans << endl;
	return 0;
}