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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;
int ans[500005],cf[500005],l[500005],r[500005];
const int bl=300;
int main(int argc, char** argv) {
	int n,m,t;
	ios::sync_with_stdio(false);
	cin >> n >> m;
	t=n/bl;
	for(int i=1;i<=n;i++)
		cin >> l[i] >> r[i];
	for(int i=1;i<=n;i++)
	{
		int lst=t;
		for(int j=bl;j>=1;j--)
		{
			int L=(l[i]-1)/j+1,R=r[i]/j;
			L=max(L,lst+1);
			if(R<L) continue;
			++cf[L],--cf[R+1];
			lst=max(lst,R);
		}
	}
	for(int i=1;i<=t;i++)
	{
		int ans=0;
		for(int j=1;j<=n;j++)
			ans+=(l[j]-1)/i!=r[j]/i;
		cout << ans << "\n";
	}
	for(int i=1;i<=m;i++)
	{
		cf[i]+=cf[i-1];
		cout << cf[i] << "\n";
	}
	return 0;
}
//dlstxdy