Codeforces Round #260 (Div. 2) C

Description

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.

Input

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).

Output

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

Examples

input

2
1 2

output

2

input

3
1 2 3

output

4

input

9
1 2 1 3 2 2 2 2 3

output

10

Note

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;
}