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,ans=0;
      vector<int>v[100006];
      priority_queue<int>q;
      cin>>n>>m;
      for(int a,b,i=0;i<n;i++){ 
          cin>>a>>b;
          v[a].push_back(b);
      } 
      for(int i=1;i<=m;i++){ 
          
          for(int j=0;j<v[i].size();j++)q.push(v[i][j]);
          if(q.empty()==false){ 
              ans+=q.top();
              q.pop();
          } 
      } 
      cout<<ans<<endl; 
      
          
      
     
        return 0;
    }