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 LuaJIT (2.0.4) (judge verdict: Time Limit Exceeded)

local mfl = math.floor
local n, m = io.read("*n", "*n")
local imos = {}
for i = 1, m + 1 do
  imos[i] = 0
end
for i_n = 1, n do
  local l, r = io.read("*n", "*n")
  if l == r then
    local lim = math.ceil(math.sqrt(r))
    for i = 1, lim do
      if r % i == 0 then
        local z = mfl(r / i)
        if i <= z then
          imos[i] = imos[i] + 1
          imos[i + 1] = imos[i + 1] - 1
          if i ~= z then
            imos[z] = imos[z] + 1
            imos[z + 1] = imos[z + 1] - 1
          end
        end
      end
    end
  else
    l = l - 1
    imos[1] = imos[1] + 1
    imos[r + 1] = imos[r + 1] - 1
    for i = 1, r do
      local rlim2 = mfl(r / (i + 1))
      local llim = mfl(l / i)
      if rlim2 < llim then
        imos[rlim2 + 1] = imos[rlim2 + 1] - 1
        imos[llim + 1] = imos[llim + 1] + 1
      else
        if llim <=  r - l then
          break
        end
      end
    end
  end
end
for i = 2, m do
  imos[i] = imos[i] + imos[i - 1]
end
imos[m + 1] = nil
print(table.concat(imos, "\n"))