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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

int main(void)
{
    int N;
    int M;
    long result = 0;

    cin >> N;
    cin >> M;

    map<int, vector<int>> bites;

    int A;
    int B;

    int MaxM = 0;

    for (int i = 0; i < N; i++)
    {
        cin >> A;
        cin >> B;
        bites[A].push_back(B);
    }

    for (int i = 1; i <=M; i++)
    {
        if (bites[i].size() > 0)
        {
            auto p = max_element(bites[i].begin(), bites[i].end());
            result += *p;
            bites[i].erase(p);
            if (i != MaxM - 1)
            {
                bites[i + 1].insert(bites[i + 1].end(), bites[i].begin(), bites[i].end());
            }
            bites[i].clear();
        }
    }

    cout << result << endl;

    return 0;
}