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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n, m = gets.split.map(&:to_i)
@ab = n.times.map {gets.split.map(&:to_i)}
@ab.sort_by!{|e| e[1]}.sort_by!{|e| e[0]}
@ab.unshift([0, 0])

x = 1 << (n - 1).bit_length
@t = Array.new(x << 1, 0)

n.times {|i| @t[x + i] = i + 1}
(x - 1).downto(1) do |i|
    @t[i] = @ab[@t[2*i]][1] > @ab[@t[2*i+1]][1] ? @t[2*i] : @t[2*i+1]
end

def max(i, s, t, l, r)
    if t < l || s > r
        0
    elsif s >= l && t <= r
        @t[i]
    else
        m = (s + t) / 2
        y = max(2*i, s, m, l, r)
        z = max(2*i+1, m+1, t, l, r)
        @ab[y][1] > @ab[z][1] ? y : z
    end
end

ans = 0
i = 0
0.upto(m) do |j|
    i += 1 while i <= n && @ab[i][0] <= j
    i -= 1

    next if i == 0
    z = max(1, 1, n, 1, i)
    next if z == 0

    ans += @ab[z][1]
    z += x - 1
    @t[z] = 0
    z /= 2
    while z != 0
        @t[z] = @ab[@t[2*z]][1] > @ab[@t[2*z+1]][1] ? @t[2*z] : @t[2*z+1]
        z /= 2
    end
end

puts ans