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 int long long
const int MAXN=2e5+69,INF=1e17;
int n;
int a[MAXN];
int dp[MAXN][3];
signed main(){
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++)for(int j=0;j<3;j++)dp[i][j]=-INF;
    dp[1][0]=a[1];
    for(int i=3;i<=n;i+=2)dp[i][0]=dp[i-2][0]+a[i];
    dp[2][1]=a[2];
    for(int i=4;i<=n;i+=2)dp[i][1]=max(dp[i-3][0],dp[i-2][1])+a[i];
    for(int i=3;i<=n;i++){
        dp[i][2]=dp[i-2][2]+a[i];
        dp[i][2]=max(dp[i][2],dp[i-3][1]+a[i]);
        if(i>=4)dp[i][2]=max(dp[i][2],dp[i-4][0]+a[i]);
    }
    int p=n&1;
    cout<<max(dp[n-1][p],dp[n][p+1]);
}