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)

#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>
using namespace std;
const int INF= 1e9+5;
typedef long long ll;
  priority_queue<ll>pq;
  vector<ll>data[100010];
int main(){
  int n,m;
  cin>>n>>m;
  vector<int>a(n);
  vector<int>b(n);
  for(int i=0;i<n;i++)cin>>a[i]>>b[i];
  for(int i=0;i<n;i++){
    data[a[i]].push_back(b[i]);
  }
  ll sum=0;
 
  for(int i=1;i<=m;i++){
    
    for(auto t:data[i]){
      pq.push(t);
    }
    if(!pq.empty()){
      sum+=pq.top();
      pq.pop();
    }
  }
  cout<<sum<<endl;
  return 0;
}