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)

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

baito = []
dp = [0] * (10**5+1)
for i in range(N):
    a, b = map(int, input().split())
    baito.append([a, b, 0])

for i in range(M):
    rewards = sorted([x for x in baito if x[0] + i <= M and x[2] == 0], key=lambda x: (-x[1], -x[0]))
    if rewards:
        max_reward = rewards[0]
        max_reward[2] = 1
        if i == 0:
            dp[i] = max_reward[1]
        else:
            dp[i] = dp[i-1] + max_reward[1]
    else:
        if i == 0:
            dp[i] = 0
        else:
            dp[i] = dp[i-1]

print(dp[M-1])