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 PAYDAY{
  int a;
  int b;
  int co;
}PD;
int cmp(const void *a,const void *b){
  return ((PD*)a)->a -((PD*)b)->a;
}
int main(void){
  long long int res=0;
  int n,m,i,j;
  scanf("%d %d",&n,&m);
  PD p[n];
  for(i=0;i<n;i++){
    scanf("%d %d",&p[i].a,&p[i].b);
    p[i].co = 0;
  }
  qsort(p,n,sizeof(struct PAYDAY),cmp);
  for(i=1;i<m+1;i++){
    int max=0;
    int tmp=-1;
    for(j=0;j<n;j++){
      if(p[j].a>i) break;
      else{
        if(max<p[j].b && p[j].co==0){
          max = p[j].b;
          tmp = j;
        }
      }
    }
    if(tmp!=-1){
      //printf("max::%d tmp::%d\n");
      res += max;
      p[tmp].co = 1;
    }
  }
  printf("%d\n",res);
  return 0;
}