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() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int N, M;
    cin >> N >> M;
    vector<pair<int, int> > AB(N);
    for (auto &ab : AB) cin >> ab.first >> ab.second;
    sort(AB.begin(), AB.end(), [](const pair<int, int> &_a, const pair<int, int> &_b) {
        return _a.second > _b.second;
    });
    int ans = 0;
    set<int> S;
    for (int i = 0; i < M; i++) {
        S.insert(i+1);
    }
    for (const auto &ab : AB) {
        if (S.empty()) break;
        auto it = lower_bound(S.begin(), S.end(), ab.first);
        if (it == S.end()) continue;
        S.erase(*it);
        ans += ab.second;
    }
    printf("%d\n", ans);
    return 0;
}