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;
const int max_n = 200005;
const ll inf = 1e18;
int n, a[max_n];
ll dp[max_n][3];
ll solve(int n, int k) {
    if(k > (n + 1) / 2) return -inf;
    if(k <= 0) return 0;
    int x = k - n/2 - 1;
    if(~ dp[n][x]) return dp[n][x];
    return dp[n][x] = max(solve(n - 1, k), solve(n - 2, k - 1) + a[n]);
}
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", a+i);
    for(int i = 0; i <= n; i++) {
        for(int j = 0; j < 3; j++) dp[i][j] = -1;
    }
    printf("%lld\n", solve(n, n/2));
    return 0;
}