Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
int main(){
  int N,M;
  cin >>N>>M;
  vector<vector<int> >data(10000000);
  for (int i=0;i<N;i++){
    int a,b;
    cin>>a>>b;
    --a;
    data.at(a).push_back(b);
  }
  for (int i=0;i<data.size();i++){
   data.at(i).resize(M);
  
  sort(data.at(i).begin(),data.at(i).end());
  reverse(data.at(i).begin(),data.at(i).end());
  }
 
  vector<int>data1;
for (int i=0;i<M;i++){
    vector<int>data2(data.at(M-1-i).begin(),data.at(M-1-i).begin()+i+1);
    data1.insert(data1.end(),data2.begin(),data2.end());
}
sort(data1.begin(),data1.end());
  reverse(data1.begin(),data1.end());
 vector<int>data3(data1.begin(),data1.begin()+M);
int p;
  for (int i=0;i<M;i++){
  p+=data3.at(i);  
  }
  cout <<p<<endl;

}