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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
using pii = pair<int, int>;

int main() {
    int n, m; cin >> n >> m;
    vector<pair<int, int>> jobs(n);
    for (auto &p: jobs) cin >> p.first >> p.second;
    sort(jobs.begin(), jobs.end(), [](pii a, pii b) {
        if (a.second == b.second) return a.first < b.first;
        else return a.second > b.second;
    });
    vector<int> can(m + 1, 0);
    for (auto p: jobs) {
        for (int i = p.first; i <= m; i++) if (can[i] == 0) {
            can[i] = p.second;
            break;
        }
    }
    long long sum = 0;
    for (int i = 0; i <= m; i++) sum += can[i];
    cout << sum << endl;
    return 0;
}