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<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
  int N, M;
  cin >> N >> M;
  priority_queue<int> option;       //残り日数に応じて実行できる仕事の報酬を追加いていく
  vector<pair<int, int>> work;     //仕事種類
  int profit = 0;                                        //報酬

  for(int i = 0; i < N; i++){            //計算量O(N)
    int t, p;
    cin >> t >> p;
    work.push_back(make_pair(t, p));
  }

  sort(work.begin(), work.end());       //時間を基準に昇順

  /*for(int i = 0; i < N; i++)
    cout << work[i].first << " " << work[i].second << endl;
  cout << "---------------------------------------" << endl;*/

  int i = 0;    //配列に格納されている仕事(日数少ない順)
  for(int t = 1; t <= M; t++){
    while(work[i].first == t){
      option.push(work[i].second);       //work[i].first日後に来る報酬を詰め込んでいく
      i++;
    }

    if(option.empty())
      continue;

    profit += option.top();                        //現状選択できる最大値を取得
    option.pop();                                            //その仕事はもう選べない  
   
  }

  cout << profit << endl;

}