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 < (n);i++)
using namespace std;
typedef long long ll;


template<typename T>
void remove(std::vector<T>& vector, unsigned int index)
{
    vector.erase(vector.begin() + index);
}


int main(){
  int n, m;
  cin >> n >> m;
  vector<vector<int>> jobs(m);
  rep(i,n){
    int a,b;
    cin >> a >> b;
    if (a > m)continue;
    jobs[m-a].push_back(b);
  }
  priority_queue<int> q;
  ll ans = 0;
  for(int i = m-1; i>= 0;--i){
    for(int b:jobs[i]){
      q.push(b);
    }
    if(!q.empty()){
      ans += q.top();
      q.pop();
    }

  }
  cout << ans << endl;
  return 0;
}