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: Wrong Answer)

#include<stdio.h>
#define N 100000

int main(void){
  int n,m,i,max,j,s,maxb,l;
  int a[N]={0};
  int b[N]={0};
  int c[N]={0};
  max=1;
  s=0;
  maxb=1;

  scanf("%d%d",&n,&m);

  for(i=1;i<=n;i++){
    scanf("%d%d",&a[i],&b[i]);
    if(max<a[i]&&a[i]<=m)
      max=a[i];
    if(a[i]>m)
      b[i]=0;
  }

  for(l=0;l<n;l++){
    for(i=1;i<=n;i++){
      if(a[i]==m-l){
	if(maxb<b[i])
	  j=i;
      }
    }
    for(i=1;i<=n;i++){
      if(a[i]==m-l&&i!=j)
	b[i]=0;
    }
  }
  for(i=1;i<=n;i++)
    s=s+b[i];

  printf("%d\n",s);
  return 0;
}