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;

int n;
ll a[200005],dp[200005][3],ans=-0x3f3f3f3f3f3f3f3fll;

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lld",a+i);
    memset(dp,0xc0,sizeof(dp));
    dp[0][1]=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<3;j++)
        {
            for(int k=i==1?1:2;k<=4;k++) if(i-k>=0)
                for(int g=0;g<3;g++) if(i/2+j-(i-k)/2-g==1)
                    dp[i][j]=max(dp[i][j],dp[i-k][g]+a[i]);
            if(i/2+j-1==n/2) ans=max(ans,dp[i][j]);
        }
    printf("%lld",ans);
    return 0;
}