Boredom

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 a ₁, a ₂, …, a _n (1 ≤ a _i ≤ 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.

Solution:

#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

const int N = 1000010;

int cnt[N];
long long f[N];

int main() {
int n;
scanf("%d", &n);
memset(cnt, 0, sizeof(cnt));
for (int i = 0; i < n; i++) {
    int foo;
    scanf("%d", &foo);
    cnt[foo]++;
  }
  f[0] = 0;
  for (int i = 1; i < N; i++) {
    f[i] = (long long)i * cnt[i];
    if (i - 2 >= 0) {
f[i] += f[i - 2];
}
if (f[i - 1] > f[i]) {
f[i] = f[i - 1];
}
}
cout << f[N - 1] << endl;
  return 0;
}