codeforces-455A

A. Boredom

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

 这道题的大概意思是输入n个数字，每次从中取出一个数k，并且将k-1和k+1抹去，然后继续取，直到取完为止，求最后取出数最大的和是多少。

想法是使用map函数，记录每个数的次数，然后从0到最大值遍历，使用动态规划的思想，每次记录两个数，一个是k-2的最大值，一个是k-1的最大值，然后用max（k-1，k-2+map[k]*k）求得当前最大值

#include<iostream>
#include<map>
#include<algorithm>
using namespace std;

map<long, int>mp;

int main()
{
    int n, num, p, q, maxn = 0;
    cin >> n;
    for (int i = 0;i < n;i++)
    {
        getchar();
        cin >> num;
        mp[num]++;
        if (num > maxn)
            maxn = num;
    }
    if (n == 1)
        cout << num << endl;
    else
    {
        p = q = 0;
        for (long long i = 1;i <= maxn;i++)
        {
            long long temp = i*mp[i] + q;
            q = max(q, p);
            p = temp;
        }
        cout << max(p,q) << endl;
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e5+5;

int n, arr, c[maxn];

int main () {
    cin >> n;
    memset(c, 0, sizeof(c));

    for (int i = 0; i < n; i++) {
        cin >> arr;
        c[arr]++;
    }

    ll p = 0, q = 0;

    for (int i = 1; i < maxn; i++) {

        ll tmp = q + 1LL * i * c[i];
        q = max(p, q);
        p = tmp;
    }

    cout << max(p, q) << endl;
    return 0;
}