CodeForces 456C - Boredom（DP）

C. 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 ita_k) and delete it, at that all elements equal toa_k + 1 anda_k - 1 also must be deleted from the sequence. That step bringsa_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

21 2

Output

2

Input

31 2 3

Output

4

Input

91 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 to2. 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 to2. In total we earn 10 points.

/*题目链接： http://codeforces.com/problemset/problem/456/C题意：给定一个含有n个整数的数组，你可以进行多次操作，每次操作从数组选一个数a[k]，然后将其删除，然后删除与a[k]-1和a[k]+1相等的数，然后可以得到a[k]分，求进行多次操作后得到的最多的分DP先将每个数*这个数的个数，得到选这个数所能得到的分然后DP,    dp[i] = max( dp[i-1]， 不选i这个数                 dp[i-2] + val[i] 选i这个数 )    要注意的是，    最终的答案不是dp[]这个数组的最后一个值    而是全部里面最大的那个HDU 2845 是这个题的升级版http://acm.hdu.edu.cn/showproblem.php?pid=2845*/#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <sstream>#include <map>#include <set>#define pi acos(-1.0)#define LL long long#define ULL unsigned long long#define inf 0x3f3f3f3f#define INF 1e18#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int, int> P;const double eps = 1e-10;const int maxn = 1e5 + 5;const int N = 1e4 + 5;const int mod = 1e8;int a[maxn];int cnt[maxn];LL val[maxn], dp[maxn];int main(void){//freopen("in.txt","r", stdin);    int n, maxx;    while (cin >> n){        memset(cnt, 0, sizeof(cnt));        memset(dp, 0, sizeof(dp));        maxx = -1;        for (int i = 1; i <= n; i++){            cin >> a[i];            cnt[a[i]]++;            maxx = max(maxx, a[i]);        }        for (int i = 1; i <= maxx; i++)            val[i] = (LL)cnt[i] * (LL)i;        memset(dp, 0, sizeof(dp));        dp[1] = val[1];        for (int i = 2; i <= maxx; i++)            dp[i] = max(dp[i-1], dp[i-2]+val[i]);        LL ans = -1;        for (int i = 1; i <= maxx; i++)            ans = max(ans, dp[i]);        cout << ans << endl;    }return 0;}