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: Wrong Answer)

import bisect

n, m = map(int, input().split())
a_list = []
for _ in range(n):
    a, b = map(int, input().split())
    if a <= m:
        a_list.append([a, b])
a_list = sorted(a_list, key=lambda x: (x[0], -x[1]))
a_list_1 = [aa[0] for aa in a_list]
a_list_2 = [aa[1] for aa in a_list]
ans = 0
for i in range(m):
    k = bisect.bisect_right(a_list_1, m - i) - 1
    a_list_1 = a_list_1[:k + 1]
    a_list_2 = a_list_2[:k + 1]
    if k >= 0:
        l = len(a_list_2) - list(reversed(a_list_2)).index(max(a_list_2)) - 1
        if l < len(a_list_2):
            if a_list_1[l] <= m - i:
                ans += a_list_2[l]
                del a_list_1[l], a_list_2[l]
    else:
        print(ans)
        exit()
print(ans)