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 <vector>
int main()
{
    int N, M;
    std::cin >> N >> M;
    std::vector<int> A(N);
    std::vector<int> B(N);
    for (int i=0; i<N; ++i) {
        std::cin >> A[i] >> B[i];
    }
    long sum = 0L;
    for (int day=M-1; day>=0; --day) {
        int max = 0;
        int maxt = -1;
        for (int t=0; t<N; ++t) {
            if (day+A[t]<=M) {
                if (max < B[t]) {
                    max = B[t];
                    maxt = t;
                // } else if (max == B[t]) {
                //    if (A[maxt] > A[t]) {
                //      maxt = t;
                // }
                }
            }
        }
        if (maxt >= 0) {
            sum += max;
            A[maxt] = M+1;
            B[maxt] = -1;
        }
    }
    std::cout << sum << std::endl;
    return 0;
}