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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
const int N=400005;
using namespace std;
typedef pair<int,int> pii;
int n,m,dp[N],cur;
vector<pii>g[N];
void add(int pos,int val){
    while(pos<N){
        dp[pos]+=val;
        pos+=pos&(-pos);
    }
}
int get(int pos){
    int ans=0;
    while(pos){
        ans+=dp[pos];
        pos-=pos&(-pos);
    }
    return ans;
}
int main()
{ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);
    cin>>n>>m;
    cur=n;
    for(int i=1;i<=n;i++){
        int l,r;cin>>l>>r;
        g[r-l+1].push_back({l,r});
    }
    for(int i=1;i<=m;i++){
        int v=cur;
        for(int j=0;j<N;j+=i)v+=get(j);
        cout<<v<<'\n';
        for(auto el:g[i]){
            cur--;
            add(el.first,1);
            add(el.second+1,-1);
        }
    }
    return 0;
}