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())
Data = [[0,0] for _ in range(N)]
for i in range(N):
    a,b = map(int,input().split())
    Data[i][0] = a
    Data[i][1] = b
Data.sort(key=lambda x:x[0])
Heap = []
heapq.heapify(Heap)

ans = 0
idx = 0
for i in range(1,M+1):
    max1 = 0
    maxindex = -1

    while idx < N:
        a = Data[idx][0]
        b = Data[idx][1]
        if a <= i:
            heapq.heappush(Heap,b*(-1))
            idx += 1
        else:
            break
    if len(Heap) > 0:
        max1 = heapq.heappop(Heap)*(-1)
        ans += max1

print(ans)