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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
  int n, m;
  scanf("%d%d", &n, &m);

  int delay[n];
  int salary[n];

  for(int job = 0; job < n; job++) {
    scanf("%d%d", &delay[job], &salary[job]);
  }

  int schedule[m];

  for(int day = 0; day < m; day++) {
    schedule[day] = -1;
  }

  for(int job = 0; job < n; job++) {

    int min_day = 0;
    for(int day = 0; day <= m - delay[job]; day++) {
      if(schedule[day] == -1) {
        min_day = day;
        break;
      }
      else {
        if(salary[schedule[min_day]] > salary[schedule[day]]) {
          min_day = day;
        }
      }
    }

    if(schedule[min_day] == -1 && min_day <= m - delay[job]) {
      schedule[min_day] = job;
    }
    else if(salary[schedule[min_day]] < salary[job]) {
      schedule[min_day] = job;
    }
  }

  long long income = 0;
  for(int day = 0; day < m; day++) {
    if(schedule[day] == -1) {
      continue;
    }
    else {
      income += (long long)salary[schedule[day]];
    }
  }

  printf("%lld\n", income);

  return 0;
}