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 <vector>
#include <algorithm>
#include <map>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    map<int, vector<int>> mm;
    for (int i = 0; i < n; ++i) {
        int a, b;
        cin >> a >> b;
        mm[b].push_back(a);
    }
    long answer = 0;
    vector<bool> d(m + 1, false);
    for (auto it = mm.rbegin(); it != mm.rend(); ++it) {
        sort(it->second.begin(), it->second.end());
        int available = it->second[0];
        while (available < m + 1 && d[available]) ++available;
        for (int dd: it->second) {
            int ddd = max(dd, available);
            while (ddd < m + 1 && d[ddd]) ++ddd;
            if (ddd < m + 1) {
                answer += it->first;
                d[ddd] = true;
                while (available < m + 1 && d[available]) ++available;
            } else {
                break;
            }
        }
    }
    cout << answer << endl;
}