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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int N,M;
  cin >> N >> M;
  vector < pair <int,int> > AB(N);
  set <int> s;
  for(int i=0;i<N;i++){
    cin >> AB.at(i).first >> AB.at(i).second;
    s.insert(AB.at(i).first);
  }
  sort(AB.begin(),AB.end());
  reverse(AB.begin(),AB.end());
  int sum=0;
  int m=M;
  int j=AB.at(0).first;
  int k=0;
  vector <int> x;
  while(true){
    if(j==0){
      break;
    }
    if((!s.count(j))||(j>M)){
      j--;
      continue;
    }
    for(int i=k;i<N;i++){
      if(AB.at(i).first==j){
        k=i;
        break;
      }
    }
    sum+=AB.at(k).second;
    m=M-1;
    for(int n=1;n<N-k;n++){
      for(int i=k+n;i<N;i++){
        if(AB.at(i).first<=m){
          sum+=AB.at(i).second;
          m--;
        }
      }
      x.push_back(sum);
      sum=AB.at(k).second;
      m=M-1;
    }
    sum=0;
    m=M;
    j--;
  }
  sort(x.begin(),x.end());
  reverse(x.begin(),x.end());
  cout << x.at(0) << endl;
  return 0;
}