codeforces-455A

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

input

3
1 2 3

output

input

9
1 2 1 3 2 2 2 2 3

output

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;
}