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: Accepted)

I, X, Y, n, m, rm, ans['   '];
main() {
  scanf("%d%d", &n, &m);
  for (; ~scanf("%d%d", &X, &Y);) {
    int b = Y + 1;
    for (int j = 1; j * j <= m; j++) {
      int tx = (X + j - 1) / j, ty = Y / j;
      ans[tx]++;
      ans[ty + 1 < b ? ty + 1 : b]--;
      b = tx;
    }
    for (I = 1; I < b; I++) {
      if ((X - 1) / I < Y / I) {
        ans[I]++;
        ans[I + 1]--;
      }
    }
  }
  for (I = 1; I <= m; I++) {
    ans[I] += ans[I - 1];
    printf("%d\n", ans[I]);
  }
}