CF 455A Boredom

A. 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 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

Copy

2
1 2

output

Copy

2

input

Copy

3
1 2 3

output

Copy

4

input

Copy

9
1 2 1 3 2 2 2 2 3

output

Copy

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】

给你一个数组a，里面有n个整数。你每次可以选择数组中的一个元素a_k，从数组中删掉它，再删掉所有值等于a_k + 1 或者 a_k - 1的元素，这样你可以得到 a_k 分。你可以重复进行多次该操作，请问你最后最多能得多少分？

【分析】

先求出数列中每一个数字k的出现次数num[k]

状态转移方程：如果取得第i-1个数，那么第i-2和第i个数均不可取；反之可取得第i和第i-2个数

第i个状态的值为 (取得第i-1个数的得分) 与 (取得第i-2个数得分和取得当前数的得分之和) 的最大值

注意，最后一重for循环要从2循环至已知的maxn

【代码】

#include<cstdio>
#include<iostream>
using namespace std;
const int N=1e5+5;
inline int read(){
	register int x=0;register char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
	return x;
}
int n,mx;long long cnt[N],f[N];
int main(){
	n=read();
	for(int i=1,x;i<=n;i++) mx=max(mx,x=read()),cnt[x]+=x;
	f[1]=cnt[1];
	for(int i=2;i<=mx;i++) f[i]=max(f[i-1],f[i-2]+cnt[i]);
	printf("%I64d\n",f[mx]);
	return 0;
}