Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Print a single integer — the maximum number of points that Alex can earn.

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this[2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

比赛中遇到的dp题，比赛时没有思路，赛后有点思路但不完善，听了讲解后算是懂了，还需要多积累。

若取当前的值，则与其相邻的值就不能取，故状态转移方程：

　　dp[i][0]=max(dp[i-1][0],dp[i-1][1]);

　　dp[i][1]=dp[i-1][0]+value[i];

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

using namespace std;

long long vis[];

long long dp[][];

long long value[];

int main()

{

    int n,num;

    scanf("%d",&n);

    for(int i=; i<=n; i++)

    {

        scanf("%d",&num);

        vis[num]++;

    }

    int cnt=;

    for(int i=;i<=1e5;i++)

    {

        dp[i][]=max(dp[i-][],dp[i-][]);

        dp[i][]=dp[i-][]+vis[i]*i;

    }

    //cout<<dp[cnt-1][0]<<endl<<dp[cnt-1][1]<<endl;

    printf("%I64d\n",dp[][]>dp[][]?dp[][]:dp[][]);

    return ;

}