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>
#define ll long long
#define INF 999999999
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;

typedef pair<int,int>P;

const int MAX_N = 100005;
const int N = 1<<17;

int bit[MAX_N+1];
vector<P> len[MAX_N];

//i番目までの和を計算する
int sum(int i)
{
	int s = 0;
	while(i>0){
		s += bit[i];
		i -= i & -i;
	}
	return s;
}
//i番目の値にxを加える
void add(int i,int x)
{
	while(i <= MAX_N){
		bit[i] += x;
		i += i & -i;
	}
}

int main()
{
	int n,m,l,r,ans;
	cin >> n >> m;
	rep(i,n){
		scanf("%d%d",&l,&r);
		len[r-l].push_back(P(l-1,r-1));
	}
	cout << n << "\n";
	int res = n;
	for(int i=1;i<m;i++){
		ans=0;
		rep(j,len[i-1].size()){
			add(len[i-1][j].first+1,1);
			add(len[i-1][j].second+2,-1);
			res--;
		}
		ans += res;
		for(int in=i+1;in<=m;in+=(i+1)){
			ans += sum(in);
		}
		cout << ans << "\n";
	}
}