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 F first
#define S second

using namespace std;
typedef long long int64;

int n;
vector<int64> a;
map<pair<int, int>, int64> dp;
int64 solve(int pos, int cnt) {
	if (pos >= n) {
		if (cnt == n / 2) return 0;
		else return -1e15;
	}
	if (cnt == n / 2) return 0;
	if ((n - pos + 1) / 2 < (n / 2 - cnt)) return -1e15;
	if (dp.find({pos, cnt}) != dp.end()) return dp[{pos, cnt}];
	return dp[{pos, cnt}] = max(a[pos] + solve(pos + 2, cnt + 1), solve(pos + 1, cnt));
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	cin >> n;
	a = vector<int64> (n);
	for (int i = 0; i < n; ++ i) {
		cin >> a[i];
	}
	cout << solve(0, 0);
}