Coderforces-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 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 brings a_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.

Example

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：就是给你N个数。你可选择大小的为A的数，但是A-1和A+1必须从数队移走。

题解：本题与数的顺序无关（这时候想到了DP），我们先对其排序。然后找其状态转移方程。假设dp[j]表示从1加到j 时最大值。则dp[j]=max(dp[j-1],dp[j-2]+i*cnt[i]).【ps:cnt[i]表示数值为】

AC代码为：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

long long  n, a[100002], cot[100002], d[100002];

int main()

{
cin >> n;

long long Max = 0;

for (int i = 1; i <= n; i++)
{
cin >> a[i];
if (a[i] > Max)
Max = a[i];
cot[a[i]]++;
}

d[1] = cot[1];  d[0] = 0;

for (int i = 2; i <= Max; i++)
d[i] = max(d[i - 1], d[i - 2] + cot[i] * i);

cout << d[Max] << endl;

return 0;

}