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 <numeric>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <map>
using namespace std;

#define maxA 100000
#define maxB 10000
int main()
{
    long long int ans = 0;
    int N, M;
    vector<vector<int>> job(maxA);
    cin >> N >> M;
    {
        int tp, tm;
        for (int i = 0; i < N; i++)
        {
            cin >> tp >> tm;
            job[tp].push_back(tm);
        }
    }
    {
        priority_queue<int> q;
        for (int i = 1; i <= M; i++)
        {
            for (auto &&j : job[i])
            {
                q.push(j);
            }
            if (q.size() != 0)
            {
                ans += q.top();
                q.pop();
            }
        }
    }
    cout << ans << endl;
    return 0;
}