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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <utility>
#include <map>
#include <algorithm>
#include <queue>
#include <cmath>

using namespace std;

typedef pair<long,long> P;

int main() {
  long n,m;
  long ans=0;
  cin >> n >> m;
  vector< P > jobs(n);
  vector< long > j(100050,0);


  for (int i=0; i<n; i++) {
    int a,b;
    cin >> a >> b;
    jobs[i] = P(a,b);
  }

  sort(jobs.begin(), jobs.end(), []( P a, P b) { return a.second > b.second; });

  for (int i=0; i<n; i++) {
    if( j[jobs[i].first] == 0 ) {
      // cout << jobs[i].first << endl;
      j[jobs[i].first] = jobs[i].second;
      continue;
    }
    int k = jobs[i].first;
    while(k<=m) {
      if( j[k] == 0 ) {
      // cout << k << endl;
      j[k] = jobs[i].second;
      continue;
      }
      k++;
    }
  }

  for (int i=0; i<=m; i++) ans += j[i];


  cout << ans << endl;
}