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;

typedef long long LL;

const int N = 3e5 + 10, M = 1e5 + 10;
int n, m;
int l[N], r[N], idx[N], len[N];
int a[M];

bool cmp(int x, int y) {
    return len[x] < len[y];
}

void add(int x, int v) {
    for (; x <= m + 1; x += x & -x) a[x] += v;
}

int query(int x) {
    int ret = 0;
    for (; x; x -= x &-x) ret += a[x];
    return ret;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) {
        scanf("%d%d", l + i, r + i); 
        len[i] = r[i] - l[i] + 1;
        idx[i] = i;
    }
    sort(idx, idx + n, cmp);
    int j = 0;
    for (int i = 1; i <= m; i++) {
        while (j < n && len[idx[j]] < i) {
            add(l[idx[j]], 1);
            add(r[idx[j]]+1, -1);
            j++;
        }
        int ans = 0;
        for (int k = i; k <= m; k += i) ans += query(k);
        printf("%d\n", ans + n - j);
    }
    return 0;
}