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;
using ll = long long;
#define rep(i,n) for (int i = 0; i < n; ++i)
#define REP(i,s,t) for (int i = s; i < t; ++i)

const int N = 2e5;
int n, a[N];
ll sum0[N+1], sum1[N+1];

int main() {
	cin >> n;
	rep(i,n) {
		cin >> a[i];
		sum0[i+1] = sum0[i];
		sum1[i+1] = sum1[i];
		if (i%2) sum1[i+1] += a[i];
		else sum0[i+1] += a[i];
	}

	ll ans = LLONG_MIN, mx = LLONG_MIN;
	if (n%2) {
		for (int i=n-1; i>=0; i-=2) {
			mx = max(mx, sum1[i]+sum0[n]-sum0[i+1]);
			ans = max(ans, sum0[i]+mx-sum1[i]);
		}
	}
	else {
		for (int i=n; i>=0; i-=2) {
			ans = max(ans, sum0[i]+sum1[n]-sum1[i]);
		}
	}

	printf("%lld\n", ans);
}