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;

int N, M;
vector<vector<int>> work(100010, vector<int>());  //work[i] : i日後にもらえる報酬一覧
priority_queue<int> pq;
int res = 0;   //報酬合計
int profit, d, p;

int main(){
  cin >> N >> M;

  for(int i = 0; i < N; i++){ cin >> d >> p; work[d].push_back(p); }

  for(int day = 1; day <= M; day++){  //残り日数
    for(int k = 0; k < work[day].size(); k++) pq.push(work[day][k]);

    if(pq.empty()) continue;  //受けれる仕事なし
    profit = pq.top();  //残り日数dayで受けれる仕事の中での最大報酬
    pq.pop();  //この仕事は受けた
    res += profit;
  }

  cout << res << endl;
  return 0;

}