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: Runtime Error)

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

int main(void){
  int n, m, i, j, k, max;
  int a[10000];
  int b[10000];
  long int money = 0;
  scanf("%d %d", &n, &m);
  for(i=0; i<n; i++)
    scanf("%d %d", a+i, b+i);
  for(i=0; i<m; i++){
    max = 0;
    k = -1;
    for(j=0; j<n; j++){
      if(max < b[j] && a[j] <= m - i){
        max = b[j];
        k = j;
      } else if(max == b[j]){
        if(a[j] <= m - i && a[k] < a[j]) k = j;
      }
    }
    if(k >= 0){
      money += max;
      b[k] = -1;
    }
  }
  printf("%ld\n", money);
  return 0;
}