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>
#define ll long long

using namespace std;

const int N = 2e5 + 5, mod = 1e9 + 7;

ll dp[N][3][3], arr[N], n;

ll solve(int ind, int jmp, int odd)
{
    if(ind >= n) return -1e17 * (jmp == 2 or odd == 1);
    ll &ret = dp[ind][jmp][odd];
    if(~ret) return ret;
    ret = solve(ind + 2, jmp, !odd) + arr[ind];
    if(jmp) ret = max(ret, solve(ind + 1, jmp - 1, odd));
    return ret;
}

int main()
{
    memset(dp, -1, sizeof dp);
    scanf("%lld", &n);
    if(n == 1) return puts("0");
    for(int i = 0 ; i < n ; i++) scanf("%lld", arr + i);
    printf("%lld\n", solve(0, 1 + (n&1), (n/2)&1));
    return 0;
}
