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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define int long
typedef struct{
	int time,money;
	bool done;
}pi;
main(){
	int n,m,res=0;
	cin>>n>>m;
	vector<pi> job(n);
	for(int i=0;i<n;++i){cin>>job[i].time>>job[i].money; job[i].done=false;}
	sort(job.begin(),job.end(),[](pi a,pi b){return a.money==b.money?a.time>b.time:a.money>b.money;});
	bool lp=true;
	int i=0,day=0;
	while(day<=m){
		if(!job[i].done&&job[i].time+day<=m){
			res+=job[i].money;
			day++;
			//cerr<<job[i].time<<","<<job[i].money<<"\n";
			job[i].done=true;
			lp=false;
		}
		if((i+1)%n==0){
			if(lp)break;
			else {lp=true; i=0;}
		}
		else i++;
	}
	cout<<res;
}