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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>

#define rep(i, n) for(int i = 0; i < n; ++i)

using namespace std;

using P = pair<int, int>;

int n, m;
int l[300000];
int r[300000];
P v[300000];
int lazy[1 << 18];
int val[1 << 18];

void add(int a, int b, int l = 0, int r = 1 << 17, int k = 0){
  if(b <= l || r <= a){
    return;
  }
  if(a <= l && r <= b){
    ++lazy[k];
    return;
  }
  int m = (l + r) / 2;
  add(a, b, l, m, k * 2 + 1);
  add(a, b, m, r, k * 2 + 2);
}
int get(int i, int l = 0, int r = 1 << 17, int k = 0){
  if(i + 1 <= l || r <= i){
    return 0;
  }
  if(i <= l && r <= i + 1){
    return val[k] + lazy[k];
  }
  val[k] += lazy[k];
  lazy[k * 2 + 1] += lazy[k];
  lazy[k * 2 + 2] += lazy[k];
  lazy[k] = 0;
  int m = (l + r) / 2;
  return get(i, l, m, k * 2 + 1) + get(i, m, r, k * 2 + 2);
}

int main(){
  cin >> n >> m;
  rep(i, n){
    cin >> l[i] >> r[i];
    v[i] = P(r[i] - l[i], i);
  }
  sort(v, v + n);
  int p = 0;
  for(int i = 1; i <= m; ++i){
    while(p < n && v[p].first < i){
      int k = v[p++].second;
      add(l[k], r[k] + 1);
    }
    int a = n - p;
    for(int j = 0; j <= m; j += i){
      a += get(j);
    }
    cout << a << endl;
  }
  return 0;
}