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: Runtime Error)

#include <bits/stdc++.h>
#define s second
using T=std::pair<int,int>;
T*P=new T[1<<19],*p;
int BIT[1<<17],j,k,l,r,x,M;
int main(){
	p=P;
	while(~scanf("%d%d",&p->first,&p->s)) p++;
	M=P->s;
	++P;
	sort(P,p,[](T a,T b){return a.s-a.first<b.s-b.first;});
	--p;
	//for(int d=0;d<p;d++)printf("%d ",P[i].first);puts("");
	while(x++<M){
		while(std::tie(l,r)=*P, l&&r-l<x){
			for(j=l; j<=M; j+=j&-j) BIT[j]++;
			for(j=r+1; j<=M; j+=j&-j) BIT[j]--;
			P++;
		}
		k=p-P;
		for(j=0; j<=M; j+=x) {
			for(l=j; l; l-=l&-l)
				k+=BIT[l];
		}
		printf("%d\n",k+1);
	}
	j=0;
}