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;
const int M=2e5+5;
const LL INF=0x3f3f3f3f3f3f3f3f;
int a[M];
map<int,LL> dp[M];
// dp[i][j] consider 1-i select j

LL getdp(int n,int k)
{
	//cout<<n<<" "<<k<<endl;
	if(k>(n+1)/2) return -INF;
	if(k==0||n<0) return 0;
	if(dp[n].count(k)) return dp[n][k];
	LL &ans=dp[n][k];
	ans=-INF;
	ans=max(getdp(n-1,k),getdp(n-2,k-1)+a[n]);
	return ans;
}

int main()
{
	int n; scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	printf("%lld\n",getdp(n,n/2));
	return 0;
}