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;

#define pi 3.1415926535
#define ll long long
#define mod 1000000007



int main() {
	ll n;
	int i;
	ll a[200010];
	ll data[200010];
	ll dp[200010];
	ll stk1, stk2;
	cin >> n;
	for (i = 1; i <= n; i++) {
		cin >> a[i];
	}
	data[1] = a[1];
	dp[0] = 0;
	dp[1] = 0;
	for (i = 3; i <= n; i+=2) {
		data[i] = data[i - 2] + a[i];
	}
	for (i = 2; i <= n; i++) {
		if (i & 1) {
			dp[i] = max(dp[i - 1], dp[i - 2] + a[i]);
		}
		else {
			dp[i] = max(dp[i - 2] + a[i], data[i - 1]);
		}
	}
	//if (n & 1) dp[n] = max(dp[n], data[n] - data[1]);;
	cout << dp[n];
	return 0;
}