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 <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>

using namespace std;

class Fenwick {
public:
	Fenwick(int _N) : N(_N), v(N+1, 0) {}
private:
	int N;
	vector<int> v;
public:
	void add(int val, int p){
		for(int i=p;i<=N;i+=i&-i) v[i] += val;
	}
	int sum(int p){
		int res = 0;
		for(int i=p;i>0;i-=i&-i) res += v[i];
		return res;
	}
};

int main(){
	std::ios_base::sync_with_stdio(false);
	int N, M;
	while(cin >> N >> M){
		Fenwick f(M+1);

		vector<pair<int,int>> r(N);
		for(int i=0;i<N;i++) cin >> r[i].first >> r[i].second;
		sort(r.begin(), r.end(), [](const pair<int,int>& a, const pair<int,int>& b){
				return a.second - a.first < b.second - b.first;
				});
		cout << N << endl;
		int c = 0;
		for(int d=2;d<=M;d++){
			while(c < N && r[c].second - r[c].first < d - 1){
				f.add(1, r[c].first);
				f.add(-1, r[c].second+1);
				++c;
			}
			int res = N - c;
			for(int j=d;j<=M;j+=d) res += f.sum(j);
			cout << res << endl;
		}
	}
}