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

21 2

Output

2

Input

31 2 3

Output

4

Input

91 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;
}