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<stdio.h>
#include<algorithm>
using namespace std;
long long dp[200001];//到第i个数取到i/2个数的最大值
long long sum[200001];//隔一个选一个的和到第i个位置的和为多少
int a[200001];
int main(){
    int n;
    int len = 1;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        sum[i] = (i>1?sum[i-2]:0)+a[i];
    }
    for(int i=2;i<=n;i++){
        if(i%2==0){
            dp[i] = max(dp[i-2]+a[i],sum[i-1]);
        }else{
            dp[i] = max(dp[i-2]+a[i],dp[i-1]);
        }
    }
    printf("%lld\n",dp[n]);
    return 0;
}
