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: Accepted)

import heapq

N,M = map(int, input().split())
a_b_list = []
for i in range(N):
    a,b = map(int, input().split())
    a_b_list.append([a,b])

a_b_list.sort()
task_list = [0] * (M+1)

heap = []
index = 0
for day in range(1, M+1):
    while index < N:
        if a_b_list[index][0] == day:
            heapq.heappush(heap, a_b_list[index][1] * -1)
            index += 1
        else:
            break
    if len(heap) != 0:
        task_list[day] = heapq.heappop(heap)

#print(task_list)
print(sum(task_list) * -1)


# for a,b in a_b_list:
# #    print(a, b)
#     if task_list[a] == 0:
# #        print("append")
#         task_list[a] = b
#     else:
# #        print("NOT append")
#         index = a + 1
#         while index <= M:
#             if task_list[index] == 0:
#                 task_list[index] = b
#                 break
#             index += 1
# print(sum(task_list))
# #print(task_list)