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<iostream>
#include<algorithm>
#include<queue>
using lint=int64_t;
using namespace std;

bool cmp(pair<lint,lint> l,pair<lint,lint> r)
{
	if(l.first!=r.first)
		return l.first>r.first;
	return l.second>r.second;
}

int main()
{
	lint N,M;
	pair<lint,lint> p[100010]{};

	cin >> N >> M;
	for(lint i=0;i<N;i++)
	{
		cin >> p[i].first;
		cin >> p[i].second;
	}

	sort(p,p+N);

	priority_queue<lint> que;
	lint index=0;
	lint ans=0;
	for(lint i=1;i<=M;i++)
	{
		while(index<N && p[index].first<=i)
		{
			que.push(p[index].second);
			index++;
		}

		if(!que.empty())
		{
			ans+=que.top();
			que.pop();
		}
	}

	cout << ans << endl;
	return 0;
}