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())
X=sorted([list(map(int, input().split())) for _ in range(N)], key=lambda x:x[0])

a=[]
ans, j = 0, 0

for i in range(1, M + 1):
  #M-1~Mから0~MまでのM日間について後ろから考える
  while (j < N) and (X[j][0] <= i):
    #i=1　1日で報酬が受け取れるものを取り出し最大の物をプラス　2日以内で報酬が…の繰り返し
    heapq.heappush(a, -X[j][1])  
    j += 1
  if len(a):
    ans += -heapq.heappop(a)  
    
print(ans)