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<math.h>
#include<stdlib.h>
 
int N,M,**G,*V,*count,*z;



int main()
{
  
  int i,j,s,t,k,u,a,b,c;
  scanf("%d %d",&N,&M);
  
  G=(int **)malloc(sizeof(int * )*N);
  for(i=0;i<N;i++)
  {
    G[i]=(int *)malloc(sizeof(int )*N);
  }
  count=(int *)malloc(sizeof(int )*(M+1));
  z=(int *)malloc(sizeof(int )*N);


  for(i=0;i<M+1;i++)
  {
    count[i]=-1;
  }
  for(i=0;i<N;i++)
  {
    z[i]=0;
  }

  
  for(i=0;i<N;i++)
  {
    scanf("%d %d",&s,&t);
    G[s][z[s]]=t;
    z[s]++;   
  }
  
  count[0]=0;
  for(i=1;i<M+1;i++)
  {
    a=0;
    k=-1;
    for(b=0;b<=i;b++)
    {
      for(c=0;c<z[b];c++)
      {
        if(G[b][c]>=a)
        {
          k=b;
          u=c;
          a=G[b][c];        
        }
      }
    }
    if(k!=-1)
    {
      G[k][u]=0;
    }
    count[i]=a;
  }

  j=0;
  for(i=1;i<M+1;i++)
  {
    j+=count[i];
  }
  printf("%d",j);

  
  return 0;
}