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 vectormax(vector<int> vec){
    sort(vec.begin(),vec.end());
    reverse(vec.begin(),vec.end());
    return vec.at(0);
}





int main(){
    int N,M;
    cin>>N;
    cin>>M;
    vector<int>A(N);
    vector<int>B(N);
    for(int i=0;i<N;i++){
        cin>>A.at(i);
        cin>>B.at(i);
    }
    
    int income;
    income=0;

    vector<int>a(N);
    vector<int>b(N);

    int saidai;
    saidai=0;

    for(int j=0;j<M;j++){
        // vector<int>a(N);
        // vector<int>b(N);
        for (int i = 0; i < N; i++){
            if(A.at(i)==j+1){
                b.at(i)=B.at(i);
            }
        }
        saidai=vectormax(b);
        income+=saidai;
        for(int k=0;k<N;k++){
            if(b.at(k)==saidai){
                b.at(k)=0;
                break;
            }
        }
        saidai=0;       
    }
    cout<<income<<endl;

}