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>
using namespace std;
using ll = long long;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<int> a(m + 2);
    for (int i = 0; i < n; i++) {
        int l, r;
        cin >> l >> r;
        int d = r - l + 1, u = 2e7;
        a[1]++;
        a[d + 1]--;
        for (int j = 1; j < 340; j++) {
            int nl = (l + j - 1) / j, nr = r / j;
            nl = max(d + 1, nl);
            nr = min(u - 1, nr);
            if (nr < nl) {
                continue;
            }
            a[nl]++;
            a[nr + 1]--;
            u = nl;
        }
        for (int k = d + 1; k < min({u, 340, m + 1}); k++) {
            if ((l + k - 1) / k <= r / k) {
                a[k]++;
                a[k + 1]--;
            }
        }
    }
    for (int i = 1; i <= m; i++) {
        cout << a[i] << endl;
        a[i + 1] += a[i];
    }
    return 0;
}