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 Lua (5.3.2) (judge verdict: Runtime Error)

local num, day = string.match(io.read(), "(%d+) (%d+)")
num, day = tonumber(num), tonumber(day)
local input = {}
for i = 1, num do
  local a, b = string.match(io.read(), "(%d+) (%d+)")
  table.insert(input, {
	day = tonumber(a),
    result = tonumber(b)
  })
end

table.sort(input, function (a, b)
    if a.result == b.result then
      return a.day >= b.day
    else
      return a.result >= b.result
    end
  end)

local count = 1
local sum = 0
while day > 0 and count <= #input do
  if day >= input[count].day then
    sum = sum + input[count].result
    day = day - 1
  end
  count = count + 1
end

print(sum)