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: Runtime Error)

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <deque>
#include <bits/stdc++.h>
//#include "testlib.h"
using namespace std;
#define ll long long
#define pii pair<int,int>
#define qi ios::sync_with_stdio(0)

bool debug=true;

/*    *************************************
	  * Written in New Computer           *
	  * The following code belongs to     *
	  * XiaoGeNintendo of HellHoleStudios *
	  *************************************
*/
template<typename T1,typename T2>ostream& operator<<(ostream& os,pair<T1,T2> ptt){
	os<<ptt.first<<","<<ptt.second;
	return os;
}
template<typename T>ostream& operator<<(ostream& os,vector<T> vt){
	os<<"{";
	for(int i=0;i<vt.size();i++){
		os<<vt[i]<<" ";
	}
	os<<"}";
	return os;
}

int n,m;
pii a[100005];

int fa[100005];

int find(int v){
//	cout<<v<<endl;
	if(fa[v]==v){
		return v;
	}
	return fa[v]=find(fa[v]);
}

void Union(int a,int b){
	a=find(a);
	b=find(b);
	if(a>b){
		fa[a]=b;
	}else{
		fa[b]=a;
	}
}

int main(int argc,char* argv[]){
	qi;
	cin>>n>>m;
	for(int i=0;i<n;i++){
		cin>>a[i].second>>a[i].first;
		a[i].second=m-a[i].second+1;
	}
	
	sort(a,a+n);
	reverse(a,a+n);
	
	for(int i=1;i<=n;i++){
		fa[i]=i;
	}
	
	int ret=0;
	
	for(int i=0;i<n;i++){
		int k=find(a[i].second);
//		cout<<a[i].first<<" "<<a[i].second<<" "<<k<<endl;
		if(k!=0){
			Union(k-1,k);
			ret+=a[i].first;
		}
	}
	
	cout<<ret;
	return 0;
}