Problem:
Score : 700 points

Problem Statement

Snuke has decided to play a game, where the player runs a railway company.
There are M+1 stations on Snuke Line, numbered 0 through M.
A train on Snuke Line stops at station 0 and every d-th station thereafter, where d is a predetermined constant for each train.
For example, if d = 3, the train stops at station 0, 3, 6, 9, and so forth.

There are N kinds of souvenirs sold in areas around Snuke Line. The i-th kind of souvenirs can be purchased when the train stops at one of the following stations: stations l_i, l_i+1, l_i+2, ..., r_i.

There are M values of d, the interval between two stops, for trains on Snuke Line: 1, 2, 3, ..., M.
For each of these M values, find the number of the kinds of souvenirs that can be purchased if one takes a train with that value of d at station 0.
Here, assume that it is not allowed to change trains.

Constraints

1 ≦ N ≦ 3 × 10^{5}

1 ≦ M ≦ 10^{5}

1 ≦ l_i ≦ r_i ≦ M

Input

The input is given from Standard Input in the following format:

N M
l_1 r_1
:
l_{N} r_{N}

Output

Print the answer in M lines. The i-th line should contain the maximum number of the kinds of souvenirs that can be purchased if one takes a train stopping every i-th station.

Sample Input 1

3 3
1 2
2 3
3 3

Sample Output 1

3
2
2

If one takes a train stopping every station, three kinds of souvenirs can be purchased: kind 1, 2 and 3.

If one takes a train stopping every second station, two kinds of souvenirs can be purchased: kind 1 and 2.

If one takes a train stopping every third station, two kinds of souvenirs can be purchased: kind 2 and 3.

Sample Input 2

7 9
1 7
5 9
5 7
5 9
1 1
6 8
3 4

Sample Output 2

7
6
6
5
4
5
5
3
2

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class BIT
    def initialize(n)
        @n = n+1
        @dat = Array.new(@n+1, 0)
    end

    def sum(i)
        s = 0
        while i >= 0 do
            s += @dat[i]
            i = (i & (i + 1)) - 1
        end
        s
    end

    def add(i, x)
        while i <= @n do
            @dat[i] += x
            i |= i + 1
        end
    end
end


N, M = gets.split.map(&:to_i)
#N, M = 3 * 10**5, 10**5

ls, rs = [], []
N.times do |i|
    ls[i], rs[i] = gets.split.map(&:to_i)
    #ls[i], rs[i] = 1, (i%M)+1
end
ls.push(1)
rs.push(M+1)

order = N.times.sort_by{|i|rs[i]-ls[i]} + [N]

bit = BIT.new(M+1)

di = order.shift
diff = rs[di]-ls[di]+1

(1..M).each do |m|
    while diff < m
        bit.add(ls[di], 1)
        bit.add(rs[di]+1, -1)
        di = order.shift
        diff = rs[di]-ls[di]+1
    end
    sum = order.length
    (1..(M/m)).each do |k|
        sum += bit.sum(k*m)
    end
    puts sum
end