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: Accepted)

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


int main() {
  
  int N, M; cin >> N >> M;
  
  vector<vector<int>> vec(100010);
  
  for (int i = 0; i < N; i++) {
    
    int A, B; cin >> A >> B;
    A--;
    vec[A].push_back(B);
  
  }
  
  int sum = 0;
  priority_queue<int> pq;
  
  for (int i = 0; i < M; i++) {
    
    if (!vec[i].empty()) {
      for (int j = 0; j < vec[i].size(); j++) {
        pq.push(vec[i][j]);
      }
    }
    
    if (!pq.empty()) {
      sum += pq.top();
      pq.pop();
    }
    
  }
  
  cout << sum << endl;
  
}