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: Time Limit Exceeded)

#include <algorithm>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#include <sstream>
#include <set>
using namespace std;
using ll = long long;

class range {private: struct I{int x;int operator*(){return x;}bool operator!=(I& lhs){return x<lhs.x;}void operator++(){++x;}};I i,n;
public:range(int n):i({0}),n({n}){}range(int i,int n):i({i}),n({n}){}I& begin(){return i;}I& end(){return n;}};

int main() {
  int N, M;
  while (scanf("%d%d", &N, &M) != EOF) {
    vector<int> sum(M + 10);

    for (int t : range(N)) {
      int lt, rt;
      scanf("%d%d", &lt, &rt);
      int lptr = lt, rptr = rt;
      for (int x = 2; ; ++x) {
        int now_lptr = (lt + x - 1) / x,
            now_rptr = rt / x;

        if (now_rptr < lptr) {
          sum[lptr] += 1;
          sum[rptr + 1] -= 1;
          lptr = now_lptr;
          rptr = now_rptr;
        } else {
          lptr = now_lptr;
        }

        if (now_lptr <= rt - lt + 1) {
          lptr = 1;
          sum[lptr] += 1;
          sum[rptr + 1] -= 1;
          break;
        }
      }
    }

    // take partial sum
    for (int i : range(M)) sum[i + 1] += sum[i];
    for (int i : range(M)) {
      printf("%d\n", sum[i + 1]);
    }
  }
}