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 {
  int day;
  int salary;
  int flg;
} job;

int cmp( const void *p, const void *q ) {
    return ((job*)p)->day - ((job*)q)->day;
}

int main(){
  int i,j,k,ans=0,N,M,tmp=0;
  scanf("%d %d",&N,&M);
  job A[N];
  for(i=0;i<N;i++){
    scanf("%d %d",&A[i].day,&A[i].salary);
    A[i].flg=0;
  }
  qsort(A,N,sizeof(job),cmp);
  for(i=1;i<=M;i++){
    j=0;
    tmp=0;
    while(A[j].day <= i){
      if((A[j].salary >= tmp) && A[j].flg==0){
        tmp=A[j].salary;
        k=j;
      }
      j++;
    }
    if(tmp > 0){
      ans+=tmp;
      A[k].flg=1;
    }
  }
  printf("%d\n",ans);
  return 0;
}