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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

N,M = map(int,input().split(" "))

A = []
B = []
for _ in range(N):
    a,b = map(int,input().split(" "))
    A.append(a)
    B.append(b)


# 日数でソート
ind_sorted = sorted(range(N), key = lambda x:A[x])

tmp_a = 0
tmp_b_ls = [0]
money = 0
for i in range(N):
    if A[ind_sorted[i]] > M:
        break
    if tmp_a != A[ind_sorted[i]]:
        d = A[ind_sorted[i]]-tmp_a
        tmp_b_ls = sorted(tmp_b_ls)[::-1]
        tmp_money = sum(tmp_b_ls[:d])
        tmp_b_ls = tmp_b_ls[d:]
        money += tmp_money
        tmp_a = A[ind_sorted[i]]
    tmp_b_ls.append(B[ind_sorted[i]])

d = M+1-tmp_a
tmp_money = sum(sorted(tmp_b_ls)[::-1][:d])
money += tmp_money

print(money)