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>
#include<iostream>
#include<cstdio>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
typedef unsigned long long ull;
typedef  long long ll;
#define INF  0x7f7f7f7f//7f7f7f7f
#define M 2000010
#define N 100010
const ll mod = 1e9 + 7;
ll a[M],dp[M],sum[M],n;
int main(void){
	scanf("%lld",&n);
	for(ll i = 1;i <= n;i++){
		scanf("%lld",&a[i]);
		if(i%2)
		sum[i] =sum[i-1]+a[i];
		else
		sum[i] =sum[i-1];
	}
	for(ll i = 2;i <= n;i++){
		if(i%2)
		dp[i] =max(dp[i-2]+a[i],dp[i-1]);
		else
		dp[i] =max(dp[i-2]+a[i],sum[i-1]);
	}
	printf("%lld\n",dp[n]);
	return 0;
}