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 Java8 (OpenJDK 1.8.0) (judge verdict: Memory Limit Exceeded)

import java.util.Scanner;

public class Main {
	static public class FenwickTree {
		int N;
		int[] t;

		public FenwickTree(final int N) {
			t = new int[N + 1];
			this.N = N;
		};

		public void add(int i, final int v) {
			if(i > N)
				return;
			i = Math.max(i, 1);
			for (int j = i; j <= t.length; j += j & -j)
				t[j - 1] += v;
		}

		public int cumsum(int i) {
			int s = 0;
			for (int j = i; 0 < j; j -= j & -j)
				s += t[j - 1];
			return s;
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.valueOf(sc.next());
		int M = Integer.valueOf(sc.next());

		FenwickTree cumSum = new FenwickTree(M);
		FenwickTree indCountSum = new FenwickTree(M);
		int[] numByLength = new int[M + 1];
		int[][] startOfSourveByLentgh = new int[M + 1][N + 1];
		
		for (int i = 1; i <= N; i++) {
			int l = Integer.valueOf(sc.next());
			int r = Integer.valueOf(sc.next());
			int length = r - l + 1;
			cumSum.add(length, 1);
			numByLength[length] += 1;
			startOfSourveByLentgh[length][numByLength[length]] = l;
		}

		sc.close();

		for (int length = 1; length <= M; length++) {
			int smallerLength = length - 1;
			for(int j = 1; j <= numByLength[smallerLength]; j++){
				int x = startOfSourveByLentgh[smallerLength][j];
				indCountSum.add(x, 1);
				indCountSum.add(x + smallerLength, -1);
			}
			int s1 = 0;
			for(int k = 0; k <= M; k += (length))
				s1 += indCountSum.cumsum(k);
			int s2 = cumSum.cumsum(M) - cumSum.cumsum(length - 1);
			System.out.println(s1 + s2);
		}
	}
}