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)

import copy
n, m = map(int, input().split())
ab_lis = []
for i in range(n):
    ab_lis.append(list(map(int, input().split())))
def rec(ab_lis, path, rew, r_day, ans):
    if r_day == 0:
        ans.append(rew)
        return
    for t_lis in ab_lis:
        if t_lis in path:
            continue
        elif t_lis[0] > r_day:
            ans.append(rew)
            continue
        path_c = copy.copy(path)
        path_c.append(t_lis)
        rec(ab_lis, path_c, rew + t_lis[1], r_day - 1, ans)   
ans = []
rec(ab_lis, [], 0, m, ans)        
print(max(ans))