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>
#define MAX 100001

int list[MAX][2];

int compare(void *a, void *b){
  if(((int*)a)[0] == ((int*)b)[0]){
    return ((int*)b)[1] - ((int*)a)[1];
  }else{
    return ((int*)a)[0] - ((int*)b)[0];
  }
}

int main(){
  int n,m,lead[MAX],
      index,
      i,j,sum=0;
  
  scanf("%d%d", &n, &m);
  for(i=0;i<n;i++){
    scanf("%d%d", &list[i][0], &list[i][1]);
  }
  
  qsort(list, n, sizeof(int) * 2, compare);
  
  i=1;
  j=0;
  while(i<=m && j<n){
    while(i < list[j][0]){
      lead[i] = -1;
      i++;
    }
    lead[i] = j;
    while(i == list[j][0]){
      j++;
      if(!(j<n)){break;}
    }
    i++;
  }
  while(i<=m){
    lead[i] = -1;
    i++;
  }
  
  for(i=1;i<=m;i++){
    index = -1;
    for(j=1;j<=i;j++){
      if(lead[j] != -1){
        if(index == -1){
          index = lead[j];
        }else if(list[lead[j]][1] > list[index][1]){
          index = lead[j];
        }
      }
    }
    if(index != -1){
      sum += list[index][1];
      if(list[index+1][0] == list[index][0]){
        lead[list[index][0]]++;
      }else{
        lead[list[index][0]] = -1;
      }
    }
  }
      
  printf("%d", sum);
  
  return 0;
}