TB. Boredom
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.
Example
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.
code
#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int N = 1e5+10,INF=0x3f3f3f3f,mod=1e9+7;
typedef pair<int,int> PII;
int T=1;
int a[N];
int f[N];
void solve(){
int n;
cin>>n;
for(int i=0;i<n;i++){
int x;
cin>>x;
a[x]++;
}
f[0]=0;
f[1]=a[1];
for(int i=2;i<=N;i++){
f[i]=max(f[i-1],f[i-2]+a[i]*i);
}
cout<<f[N]<<endl;
}
signed main(){
// cin>>T;
while(T--){
solve();
}
return 0;
}