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;
int a[200010];
long long f[200010];
long long g[200010];
long long get(int x) {
    if (x <= 0) return 0;
    return f[x];
}
int main() {
    int n, i;
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        if (i) g[i] = g[i-1];
        if (i%2==0) g[i]+= a[i];
    }
    for (i = 0; i < n; i++) {
        if ((i+1)%2==0) {
            f[i]=max(a[i]+get(i-2),g[i-1]);
        } else {
            f[i]=max(a[i]+get(i-2),f[i-1]);
        }
    }
    printf("%lld\n", f[n-1]);
    return 0;
}
