Codeforces Round #260 (Div. 1) Boredom（DP）

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.

【题意】给你一个序列，现在要将所有的数删除。如果删除了a[k],则a[k]+1和a[k]-1都将被删除，此次删除的收益为a[k]。求最大收益。

【分析】DP。我们将数字1~maxn依次遍历，dp[i]表示当前数字获得的最大收益，则有两种情况。一：i这个数字在数组中没有，则

dp[i]=i-2<0?0:dp[i-2];二：这个数字在数组中存在，则他可能从i-2的数字那遍历过来，也有可能从i-3的地方遍历过来，继续更新就行了，然后在删除最后一个或者倒数第二个的时候取一下最大值就行了。

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int N = 1e5+;
int n,maxn;
int a[N],cnt[N];
LL dp[N];
int main(){
    scanf("%d",&n);
    for(int i=,x;i<=n;i++){
        scanf("%d",&a[i]);
        maxn=max(maxn,a[i]);
        cnt[a[i]]++;
    }
    LL ans=;
    for(int i=;i<=maxn;++i){
        if(!cnt[i]){
            dp[i]=i-<?:dp[i-];
            if(i>=maxn-)ans=max(ans,dp[i]);
            continue;
        }
        dp[i]=1LL*((i-<?:dp[i-])+1LL*cnt[i]*i);
        dp[i]=max(dp[i],1LL*((i-<?:dp[i-])+1LL*cnt[i]*i));
        //printf("i:%d dpi:%lld\n",i,dp[i]);
        if(i>=maxn-)ans=max(ans,dp[i]);
    }
    printf("%lld\n",ans);
    return ;
}