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



Sample input:
3 4
4 3
4 1
2 2


Sample output:
5

Submission status: Runtime Error

Solution code:
n,m = map(int,input().split())
a = []
b = []
d = 100
sum = 0
for i in range(n):
  c = list(map(int,input().split()))
  if m >= c[0]:
    a.append(c[0])
    b.append(c[1])
if not b:
  print(0)
  exit()

for i in range(m):
  for j in range(n):
    f = max(b)
    if b[j] == f:
      if d > a[j]:
        d = b[j]
        e = j
  #print(e)
  a[e] = 0
  b[e] = 0
  #print(a,b)
  sum += d
  d = 100
  for k in range(n):
    if a[k] > m - i - 1:
      a[k] = 0
      b[k] = 0
print(sum)