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())

CBList = []

for i in range(N):
    A, B = map(int, input().split())
    C = M - A
    CBList.append((C, B))

CBList = sorted(CBList)

ansList = [0] * (M+1)
for i in range(N):
    day = CBList[i][0]
    if day > 0:
        minB = min(ansList[0:day+1])
        minday = ansList.index(minB)
        if minB < CBList[i][1]:
            ansList[minday] = CBList[i][1]
    elif day == 0 and ansList[0] < CBList[i][1]:
        ansList[0] = CBList[i][1]
print(sum(ansList))