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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
int n,m,l[100005];
long long ans;
struct node{int a,b;}c[100005];
int L(int x){
	if(x==l[x])return l[x];
	return l[x]=L(l[x]);
}
int cmp(node x,node y){
	return x.b>y.b;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<=m;i++)l[i]=i;
	for(int i=1;i<=n;i++)scanf("%d%d",&c[i].a,&c[i].b);
	sort(c+1,c+n+1,cmp);
	for(int i=1;i<=n;i++){
		int j=m-c[i].a+1;
		if(j<1)continue;
		l[j]=L(l[j]);j=l[j];
		if(j>0){ans+=c[i].b;l[j]--;}
	}
	printf("%lld",ans);
}