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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

while True:
    try:
        N, M = map(int, input().split())
        jobs = dict()
        for i in range(N):
            A, B = map(int, input().split())
            if A not in jobs:
                jobs[A] = list()
            jobs[A].append(B)
        result = 0
        jobs_keys = sorted(jobs.keys())
        for j in range(M):
            m = m_s = None
            for s in jobs_keys:
                if s > j+1:
                    break
                if len(jobs[s]) == 0:
                    continue
                _m = max(jobs[s])
                if m is None:
                    m = _m
                    m_s = s
                    continue
                if _m < m:
                    continue
                if _m > m or s > m_s:
                    m = _m
                    m_s = s
            if m is not None:
                jobs[m_s].remove(m)
                result += m
        print(result)
    except EOFError:
        break