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;

signed main () {
	int n, m;
	cin >> n >> m;
	vector < pair < int, int > > oneoff(n);
	for (int i = 0; i < n; ++i) 
		cin >> oneoff[i].second >> oneoff[i].first;
	sort(oneoff.rbegin(), oneoff.rend());
	set < int > days;
	for (int i = 0; i <= m; ++i)
		days.insert(i);
	int64_t ans = 0;
	for (int i = 0; i < n; ++i) {
		int a = oneoff[i].second, b = oneoff[i].first;
		if (a <= m) {
			auto j = days.upper_bound(m - a);
			if (!days.empty() && j != days.begin()) {
				days.erase(prev(j));
				ans += b;
			}
		}
	}
	cout << ans << '\n';
}