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>
long long n,a,f[(int)2e5+10],s;
int main(){
    scanf("%lld",&n);
    for(int i = 1; i <= n; ++i){
        scanf("%lld",&a);
        if(i&1)s+=a; // s is sum of odd indices
        if(i == 1)continue;
        if(i&1)f[i] = std::max(f[i-2]+a,f[i-1]); // at odd place either f[i-2]+a (common sense) or previous
        else f[i] = std::max(f[i-2]+a,s); // at even either normal or we just take odd ones
    }
    return printf("%lld\n",f[n]),0;
}