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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

typedef struct parttime{
  int day;
  int money;
}pt;

int cmp(const void *x, const void *y) {
  if (((pt *)x)->money - ((pt *)y)->money == 0) {
    return ((pt *)x)->day - ((pt *)y)->day;
  }
  return ((pt *)y)->money - ((pt *)x)->money;
}

int main() {
  int n, m, ans = 0;
  scanf("%d%d", &n, &m);
  pt c[n]; int d[m];
  for (size_t i = 0; i < m; i++) {
    d[i] = 0;
  }
  for (size_t i = 0; i < n; i++) {
    scanf("%d%d", &(c+i)->day, &(c+i)->money);
  }
  qsort(c, n, sizeof(pt), cmp);
  for (size_t i = 0; i < n; i++) {
    if (m-(c[i].day) >= 0) {
      for (int j = m-(c[i].day); j >= 0; j--) {
        if (d[j] == 0) {
          d[j] = i+1;
          ans += c[i].money;
          break;
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}