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)

from collections import Counter

n, m = map(int, input().split())
ab = sorted([list(map(int, input().split())) for i in range(n)])
c = [[]] * m
a, b = ab[0]
last = a
ad = [b]
for i in range(1, n):
    a, b = ab[i]
    if a == last:
        ad.append(b)
    else:
        if last <= m:
            c[last - 1] = ad
        ad = [b]
    last = a
if a <= m:
    c[a - 1] = ad
choice = Counter()
ans = 0
for i in range(m):
    #choice += Counter(c[i])
    if choice:
        use = max(choice)
        ans += use
        choice[use] -= 1
print(ans)