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"
using namespace std;


int main() {
	int N;
	cin >> N;
	vector<long long> A(N);
	for (int n = 0;n<N;++n) {
		cin >> A[n];
	}
	vector<vector<long long>> DP(N, vector<long long >(2, -1e18));
	// i 番目までからj個を選んだ
	// jの範囲は i/2 -1 から(i+1)/2)
	for (int i = 1; i <= N; ++i) {
		for (int j = (i + 1) / 2; j >= i / 2 - 1; --j) {
			if (j < 0) {
				continue;
			}
			DP[j][0] = max(DP[j][0] ,DP[j][1]); 
			if (j>0) {
				if (-1e18 != DP[j - 1][0]) {
					DP[j][1] = DP[j - 1][0] + A[i-1];
				}
				else {
					DP[j][1] = A[i-1];
				}
			}
		}
	}
	long long ans = 0;
	ans = max(DP[N/2][0], DP[N/2][1]);
	cout << ans << endl;
	return 0;
}

