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>
typedef long long ll;
const int maxn=2e5+9;
using namespace std;
int main()
{
	ll n,a[maxn]={0},ans;
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	ll dp[maxn][3]={0},k=n/2;
	if(n&1)
	{
		for(int i=1;i<=k;i++)
		{
			dp[i][0]=dp[i-1][0]+a[i*2-1];
			dp[i][1]=max(dp[i-1][0],dp[i-1][1])+a[i*2];
			dp[i][2]=max(dp[i-1][0],max(dp[i-1][2],dp[i-1][1]))+a[i*2+1];
		}
		ans=max(max(dp[k][0],dp[k][1]),dp[k][2]);	
	}
	else
	{
		for(int i=1;i<=k;i++)
		{
			dp[i][0]=dp[i-1][0]+a[i*2-1];
			dp[i][1]=max(dp[i-1][0],dp[i-1][1])+a[i*2];
		}
		ans=max(dp[k][0],dp[k][1]);
	}
	printf("%lld\n",ans);
	
	
}