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 <vector>
#include <math.h>
#include <string>
#include <ios>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <queue>

#define LL long long
using namespace std;
int main() {
	LL N,M;
	LL res = 0;
	cin >> N >> M;
	map<int, priority_queue<int>> salary;

	for (int i = 0; i < N; i++) {
		LL A, B;
		cin >> A >> B;
		salary[A].push(B);
	}
	
	for (int max_m = 1; max_m <= M; max_m++) {
		LL max_salary = 0;
		LL max_salary_idx = -1;
		for (int m = 1; m <= max_m; m++) {
			if (salary[m].size() > 0) {
				if (max_salary < salary[m].top()) {
					max_salary_idx = m;
				}
			}
		}

		if (max_salary_idx >= 0) {
			res += salary[max_salary_idx].top();
			salary[max_salary_idx].pop();
		}
	}
	cout << res;

	return 0;
}