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<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
ll dp[200010][5];
const ll INF = -1e16;
ll a[200010],n;
ll func(ll here, ll ch) {
	if (here >= n)
	{
		if (ch == 0)
			return 0;
		else
			return INF;
	}
	if (ch < 0)return INF;
	ll& ret = dp[here][ch];
	if (ret != -1)return ret;
	ret = INF;

	ret = max(ret, func(here + 2, ch) + a[here]);
	ret = max(ret, func(here + 1, ch - 1));
	return ret;
}
int main()
{
	memset(dp, -1, sizeof(dp));
	scanf("%lld", &n);
	for (int i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	ll sum = 0;
	sum = n / 2;
	sum += sum;
	if (n % 2)
		a[n] = -1e10;
	n++;
	printf("%lld", func(0, n-sum));
}