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 <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN=2e5+10;
int N;
int arr[MAXN];

typedef long long int ll;

ll dp[MAXN][3];

int main(){
    scanf("%d",&N);
    for(int i=1;i<=N;i++) scanf("%d",arr+i);

    dp[1][0]=0;
    dp[1][1]=arr[1];
    dp[0][0]=0;
    dp[2][0]=0;

    for(int i=0;i<=N;i++){
        for(int j=max(i/2-1,0);j<=(i+1)/2;j++){
            // printf("i=%d j=%d\n",i,j);
            if (i+2<=N) dp[i+2][(j+1)%3]=dp[i][j%3]+arr[i+2];
            if (i+1<=N) dp[i+1][j%3]=max(dp[i+1][j%3],dp[i][j%3]);
        }
    }

    printf("%lld\n",dp[N][(N/2)%3]);

    return 0;
}