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 <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <set>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
vector<pair<int, int>> A(100010);

int main(){
    int N,M;
    cin >> N >> M;
    int q,w;
    rep(i,N){
        cin >> q >> w;
        A[i] = make_pair(q,w);
    }
    sort(A.begin(),A.end(),
    [](const pair<int,int>& x, const pair<int,int>& y) {
        if(x.second > y.second){
            return true;
        }else if(x.second == y.second){
            return (x.first < y.first);
        }else{
            return false;
        }
    });
    int cu = 1;
    int cnt=0;
    //rep(j,N){
      //  cout << A[j].first <<"::"<<A[j].second << endl;
    //}
    int flag[100010]={};
    while(M >= cu){
        rep(i,N){
            if(A[i].first <= cu && flag[i] == 0){
                cnt += A[i].second;
                //A.erase(A.begin()+i);
                flag[i] = 1;
                //N--;
                //rep(j,N){
                //    cout << A[j].first <<","<<A[j].second << endl;
                //}
                //cout << endl;
                break;
            }
        }
        cu++;
    }

    cout << cnt << endl;
}