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>


int main()
{
	int t,n,m,i,x,j;
	long long int sum;
	scanf("%d %d",&n,&m);
	int a[n],b[n];
	for(i=0;i<n;i++)
		scanf("%d %d",&a[i],&b[i]);
	
	
	for(i=0;i<n-1;i++)
	{
		for(j=i+1;j<n;j++)
		{
			if(a[i]>a[j])
			{
				t=a[j];
				a[j]=a[i];
				a[i]=t;
				
				t=b[j];
				b[j]=b[i];
				b[i]=t;
			}
		}
	}
	sum=0;
	for(i=m;i>0;i--)
	{
		x=m+1-i;
		t=-1;
		for(j=0;j<n;j++)
		{
			if(a[j]>x)
				break;
			if(t==-1 && a[j]!=0)
				t=j;
			else if((b[j]==b[t]) && (a[j]>a[t]) && a[j]!=0)
				t=j;
			else if((b[j]>b[t]))
				t=j;
		}
		
		sum=sum+b[t];
		a[t]=b[t]=0;
	}
	printf("\n%lld",sum);
	return 0;
}