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: Runtime Error)

# coding: utf-8

if __name__ == "__main__":
    n, m = map(int, input().split(" "))
    ab = {}
    for i in range(n):
        a, b = map(int, input().split(" "))
        try:
            ab[a].append(b)
        except:
            ab[a] =[b]

    ans = 0
    for today in range(m):
        days = m - today
        l = {}
        for i in range(days+1)[::-1]:
            try:
                l[i] = max(ab[i])
            except:
                pass
        max_item = max(l.items(), key=lambda x: x[1])
        ans += max_item[1]
        ab[max_item[0]].remove(max_item[1])
    print(ans)