（CodeForces

（CodeForces - 455A）Boredom

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

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 ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak 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, …, an (1 ≤ ai ≤ 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.

题目大意：一个有n个数的数字串，当你将这个数字串中的ak这个数删除时，这个数字串中所有的ak-1和ak+1都将会被删除（注意这不是数组下标，即不是这个数的相邻两个数，因为这样wa了几发），相应的你会得到ak分数，问当你把所有数字都删除的时候所能获得的最大分数是多少。

思路：可从dp角度考虑。设f[i]表示i这个数字删或者不删时所能得到分数的最大值。当i这个数字不删时，那么必定i-1会删除，此时f[i]=f[i-1]；当i这个数字删除时，则f[i]=f[i-2]+i*num[i]（num[i]记录数字i出现的个数）。
综上所述，f[i]=max(f[i-1],f[i-2]+i*num[i])。
（其中初值为f[0]=0,f[1]=num[1])
显然答案是，f[mx] (mx表示数字串中最大的那个数）。
ps：最大的情况是105∗105会爆int，所以要开long long

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;const int maxn=100005;LL f[maxn],num[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int x,mx=0;        memset(num,0,sizeof(num));        for(int i=0;i<n;i++)        {            scanf("%d",&x);            if(mx<x) mx=x;            num[x]++;        }        f[0]=0;        f[1]=num[1];        for(int i=2;i<=mx;i++)            f[i]=max(f[i-1],f[i-2]+i*num[i]);        printf("%lld\n",f[mx]);    }    return 0;}