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 Int int64_t

using namespace std;

int main() {
	constexpr Int INF = 1e18;
	int N;
	cin >> N;
	vector<Int> a(N);
	for (int i = 0; i < N; ++i) { cin >> a[i]; }

	int skip = 1 + N % 2;
	vector<vector<Int>> dp(N + 2, vector<Int>(skip + 1, -INF));
	dp[0][0] = 0;
	for (int i = 0; i <= N; ++i) {
		for (int j = 0; j <= skip; ++j) {
			if (j < skip) {
				dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
			}
			if (i < N) {
				dp[i + 2][j] = max(dp[i + 2][j], dp[i][j] + a[i]);
			}
		}
	}
	cout << dp[N + 1][skip] << endl;

	return 0;
}
