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：

　　每次从n个数中选中数值为A_k将其删去，并将值为A_k-1的数和值为A_k+1的数全部删去，得到A_k的分数。问最大可获得的分数。

题解：

　　DP一下就好了。

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int maxn=1e5+5;long long cnt[maxn],dp[maxn];int main(){    int n;    while(cin>>n)    {        memset(cnt,0,sizeof(cnt));        for(int i=0;i<n;i++)        {            int x;            cin>>x;            cnt[x]++;        }        dp[0]=0,dp[1]=cnt[1];        long long ans=0;        for(int i=2;i<maxn;i++)        {            dp[i]=max(dp[i-1],dp[i-2]+cnt[i]*i);            ans=max(ans,dp[i]);        }        cout<<ans<<endl;    }    return 0;}