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 <bits/stdc++.h>
using namespace std;
#define ls l,mid,now<<1
#define rs mid+1,r,now<<1|1
const int N=310000,M=110000;
int n,m,cnt;
struct node
{
	int l,r;
	friend bool operator < (const node &r1,const node &r2)
	{return r1.r<r2.r;}
}a[N];
vector<int>vec[M];
int b[M*20],pos[M],ed[M],beg[M],nv[M*20];
int val[M*20],bj[M*80],ans[M];
void pushdown(int now)
{
	bj[now<<1]+=bj[now];
	bj[now<<1|1]+=bj[now];
	bj[now]=0;
}
void modify(int l,int r,int now,int pos,int v)
{
	if(l==r)
	{
		val[l]+=bj[now]*nv[l];
		bj[now]=0;nv[l]=v;
		return;
	}
	int mid=(l+r)>>1;
	pushdown(now);
	if(mid>=pos)modify(ls,pos,v);
	else modify(rs,pos,v);
}
void insert(int l,int r,int now,int lq,int rq)
{
	if(lq<=l&&r<=rq)
		{bj[now]++;return;}
	int mid=(l+r)>>1;
	if(mid>=lq)insert(ls,lq,rq);
	if(mid<rq) insert(rs,lq,rq);
}
void down(int l,int r,int now)
{
	if(l==r)
	{
		val[l]+=bj[now]*nv[l];
		return;
	}
	int mid=(l+r)>>1;
	pushdown(now);
	down(ls);down(rs);
}
int main()
{
	//freopen("tt.in","r",stdin);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].l,&a[i].r);
	sort(a+1,a+1+n);
	for(int i=1;i<=m;i++)
		for(int j=i;j<=m;j+=i)
			vec[j].push_back(i);
	for(int i=1;i<=m;i++)
	{
		beg[i]=cnt+1;
		for(int j=0;j<vec[i].size();j++)
			b[++cnt]=vec[i][j];
		ed[i]=cnt;
	}
	for(int i=1,now=1,t=0;i<=m;i++)
	{
		while(t<ed[i])
		{
			t++;
			if(pos[b[t]])
				modify(1,cnt,1,pos[b[t]],0);
			modify(1,cnt,1,t,1);
			pos[b[t]]=t;
		}
		while(now<=n&&a[now].r==i)
		{
			insert(1,cnt,1,beg[a[now].l],ed[a[now].r]);
			now++;	
		}
	}
	down(1,cnt,1);
	for(int i=1;i<=cnt;i++)
		ans[b[i]]+=val[i];
	for(int i=1;i<=m;i++)
		printf("%d\n",ans[i]);
	return 0;
}