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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

typedef long long ll;
typedef struct pair<int,int> P;

int a[100001]={};

vector<int> v[100001];
int n,m;


int main() {
	cin>>n>>m;
	int l,r;
	for(int i=1;i<n+1;i++){
		cin>>l>>r;
		a[l]++;
		a[r+1]--;
		v[r].push_back(l);
	}
	for(int i=1;i<m+1;++i)a[i]+=a[i-1];


	for(int i=1;i<m+1;++i){
		if(v[i].empty())continue;
		for(int j=1;j*j<=i;++j){
			int k1=i%j;
			int j2=i/j;
			for(auto it:v[i]){
				if(i-it>=k1){
					if(j<it)a[j]++;
					if(j2<it)a[j2]++;
				}
			}
		}
	}
	for(int i=1;i<m+1;++i)	cout<<a[i]<<endl;
	return 0;
}