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 <vector>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
const int N = 2e5+7,INF = 0x7f7f7f7f;
ll a[N],f[N],s[N];
int main()
{
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;++i)	scanf("%lld",&a[i]);
	s[1] = a[1];
	for(int i = 2;i <= n;++i)
		s[i] = s[i - 2] + a[i];
	f[2] = max(a[1],a[2]);
	for(int i = 3;i <= n;++i)
		if(i & 1)	f[i] = max(f[i - 1],f[i - 2] + a[i]);
		else	f[i] = max(f[i - 2] + a[i],s[i - 1]);
    printf("%lld",f[n]);
    return 0;
}
