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>
#define int long long
using namespace std;

signed main(){
	int N,M; cin>>N>>M;
	vector<pair<int,int>> AB(N);
	for(int i=0;i<N;i++){
		cin>>AB[i].first>>AB[i].second;
	}
	sort(AB.begin(),AB.end());
	int p=0,ans=0;
	priority_queue<pair<int,int>> que;
	for(int i=1;i<=M;i++){
		while(AB[p].first<=i and p<N){
			pair<int,int> tmp=make_pair(AB[p].second,AB[p].first);
			que.push(tmp);
			p++;
		}
		if(not que.empty()){
			pair<int,int> tmp=que.top();
			que.pop();
			ans+=tmp.first;
		}
	}
	cout<<ans<<endl;
	return 0;

}