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)

from heapq import heapify, heappush, heappop

n, m = map(int, input().split())

day_value_dict = {}
for _ in range(n):
    a, b = map(int, input().split())
    day_value_dict[a] = day_value_dict.get(a, [])
    heappush(day_value_dict[a], -b)


res = 0
for m_i in range(1, m + 1):
    temp = 0
    selected_x = 0
    for x in range(1, m_i + 1):
        if x not in day_value_dict or not day_value_dict[x]:
            continue
        if day_value_dict[x][0] < temp:
            temp = day_value_dict[x][0]
            selected_x = x
    if selected_x:
        res += temp
        heappop(day_value_dict[selected_x])

print(-res)