B

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.

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.

题意：选择一个数ak删除,并且删除ak+1和ak-1，同时得到ak分，求最大可以获得的分

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
ll a[maxn],b[maxn];
ll dp[maxn];
int main()
{
    ll n,m=0;
    scanf("%lld",&n);
    memset(dp,0,sizeof(dp));
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        if(m<a[i])
          m=a[i];
        b[a[i]]++;
    }
    dp[0]=0;
    dp[1]=b[1];
    for(int i=2;i<=m;i++)
    {
        dp[i]=max(dp[i-1],dp[i-2]+b[i]*i);
    }
     cout<<dp[m]<<endl;
     return 0; 
} 
//dp[i]=max(dp[i-2]+n[i]*i，dp[i-1])
//dp[i]表示的是前i个的最大值，对于第i个有取和不取的情况，
//对于可以取到i的情况，删掉的一定是i-1这个点，
//剩下的就是前i-2的情况了，由于dp[i-2]包括了取i-2的情况，于是就不用再重复考虑i-2也要加一遍的情况了，然
//后再考虑不取i的情况，就不用考虑i了，于是就是dp[i-1]了