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.

 题意：我们每次都删除一个数字n，并且n-1，n+1也删除，我们要获得n的总和要最大

解法：DP，我们记录每个数字出现次数，这样删除之后还能乘以i得到一个和，我们还要把最大的数字找出来

这样我们就从2—>max 有dp[i]=dp[i-1]，因为i-1也要删除的，dp[i]=dp[i-2]+i*出现次数，求他们之间的最大值

注意起始dp[0],dp[1]

#include<bits/stdc++.h>using namespace std;map<long long,long long>q;long long dp[100010];int main(){    long long n;    long long m;    long long MAX=-1;    cin>>n;    for(int i=1;i<=n;i++)    {        cin>>m;        q[m]++;        MAX=max(MAX,m);    }    dp[0]=0;    dp[1]=q[1];    for(int i=2;i<=MAX;i++)    {        dp[i]=max(dp[i-1],(long long)(dp[i-2]+i*q[i]));    }    cout<<dp[MAX]<<endl;    return 0;}