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 <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long ll;

bool dt[200000][3];
ll dp[200000][3];
int n;
ll A[200000];

ll f(int i, int j) {
    if(i >= n) return (n & 1) * (j ? 0 : -1e10);
    if(dt[i][j]) return dp[i][j];
    dt[i][j] = 1;
    if(j == 2) return dp[i][j] = A[i] + f(i + 2, 2);
    return dp[i][j] = max(A[i] + f(i + 2, j), f(i + 1, j + 1));
}

int main() {
    cin >> n;
    memset(dt, 0, sizeof dt);
    for(int i = 0; i < n; ++i)
        cin >> A[i];
        
    cout << f(0, n % 2 == 0);
    
    return 0;
}