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>

void input_num(long arr[],long a,long b,long m){
	long i;

	for(i=a;i<=m;i++){
		if(b>arr[i]){
			input_num(arr,i+1,arr[i],m);
			arr[i]=b;
			break;
		}
	}
}

int main(void){
  long i,j;

  long n,m;
  scanf("%ld%ld",&n,&m);

  long a,b,ans[m+1];
	for(i=0;i<=m;i++){
		ans[i]=0;
	}
  for(i=0;i<n;i++){
    scanf("%ld%ld",&a,&b);
		input_num(ans,a,b,m);
  }

	long sum=0;
	for(i=1;i<=m;i++){
		sum+=ans[i];
	}

  printf("%ld\n",sum);

  return 0;
}