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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#include <iostream>
#include <ctype.h>
#include <string>
typedef long long ll;
using namespace std;

int main(){
  int n,m;
  cin>>n>>m;
  vector <pair<int, int >> pair(n);
  rep(i,n){
    cin>>pair[i].first>>pair[i].second;
  }
    
  sort(pair.begin(), pair.end());   
  //cout<<pair[0].second;
  ll sum=0;     
  priority_queue<int> salary;
  int j=0;
  for(int i=1; i<m+1; i++){
    for( j; pair[j].first<=i; j++){
      salary.push(pair[j].second);
    }
    if(salary.empty()) continue;
    else{
      sum+=salary.top();
      salary.pop();
    }
  }
  
  cout<<sum;
}