codeforce 455A—— DP—— Boredom

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 a_k) and delete it, at that all elements equal to a_k + 1 and a_k - 1 also must be deleted from the sequence. That step bringsa_k 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, ..., a_n (1 ≤ a_i ≤ 10⁵).

Output

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

Sample Input

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

Hint

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.

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

long long  dp[100010][2];

int a[100010];

int temp[100010];

int n;

int main()

{

      while(~scanf("%d", &n)){

          memset(dp, 0, sizeof(dp));

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

            scanf("%d", &a[i]);

            temp[a[i]]++;

        }



        dp[1][1] = temp[1];

        dp[1][0] = 0;

        for(int i = 2; i <= 100000; i++){

            dp[i][1] = dp[i-1][0] + 1ll*temp[i]*i;

            dp[i][0] = max(dp[i-1][1], dp[i-1][0]);

        }



        printf("%lld\n", max(dp[100000][1], dp[100000][0]));

    }

    return 0;

}