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())
AB = []
for i in range(N) :
    AB.append(list(map(int, input().split())))

AB = sorted(AB, key=lambda x:(x[0],x[1]))

ans = 0
#print(AB)
for i in range(1, M + 1) :
    maxIncome = 0
    anker = -1
    for j in range(len(AB)) :
        #print(AB[j][0])
        if AB[j][0] <= i :
            if AB[j][1] > maxIncome :
                maxIncome = AB[j][1]
                anker = j
                #print('A', j, maxIncome)
                continue
        else :
            break
    ans += maxIncome
    if len(AB) == 0 :
        break
    if anker != -1:
        AB.pop(anker)

print(ans)