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 <algorithm>
#include <vector>

using namespace std;

struct AB{ int A, B;bool used;};

int main(void){
    int N, M, a, b;
    cin >> N >> M;
    
    vector<AB> data(N);
    for(int n=0; n<N; n++){
        cin >> a >> b;
        data[n] = AB{a,b, false};
    }
    sort(data.begin(), data.end(), [](const AB &x, const AB &y){return x.B > y.B;});
    
    long long ans = 0;
    for(int restday = 1; restday <= M; restday++){
        for(int i=0; i<N; i++){
            if(!data[i].used && data[i].A <= restday){
                ans += data[i].B;
                data[i].used = true;
                break;
            }
        }
    }
    cout << ans << endl;
}