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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=1<<17,INF=1<<30;

int n;
ll dat[2*MAX-1],lazy[2*MAX-1];

void init(int n_){
    n=1;
    while(n<n_) n*=2;
    for(int i=0;i<2*n-1;i++){
        dat[i]=0;
        lazy[i]=0;
    }
}

void eval(ll k,int l,int r){
    
    if(lazy[k]!=0){
        dat[k]+=lazy[k];
        
        if(r-l>1){
            lazy[k*2+1]+=lazy[k]/2;
            lazy[k*2+2]+=lazy[k]/2;
        }
        
        lazy[k]=0;
    }
}

void add(int a,int b,ll x,int k,int l,int r){
    eval(k,l,r);
    
    if(r<=a||b<=l) return;
    
    if(a<=l&&r<=b){
        lazy[k]+=(r-l)*x;
        eval(k,l,r);
    }
    
    else{
        add(a,b,x,2*k+1,l,(l+r)/2);
        add(a,b,x,2*k+2,(l+r)/2,r);
        dat[k]=dat[2*k+1]+dat[2*k+2];
    }
}

ll query(int a,int b,int k,int l,int r){
    
    if(r<=a||b<=l) return 0;
    
    eval(k,l,r);
    
    if(a<=l&&r<=b) return dat[k];
    
    ll vl=query(a,b,2*k+1,l,(l+r)/2);
    ll vr=query(a,b,2*k+2,(l+r)/2,r);
    return vl+vr;
}


int main(){
    
    int N,M;cin>>N>>M;
    
    vector<int> cnt(M+3,0);
    
    init(M+3);
    
    for(int i=0;i<N;i++){
        int a,b;
        scanf("%d",&a);
        scanf("%d",&a);
        if(a==b){
            for(int j=1;j*j<=a;j++){
                if(j*j==a) add(j,j+1,1,0,0,n);
                else if(a%j==0){
                    add(j,j+1,1,0,0,n);
                    add(a/j,a/j+1,1,0,0,n);
                }
            }
        }else if(a-1<=b-a+1) add(1,b+1,1,0,0,n);
        else if(b-a+1<=20){
            set<int> S;
            for(int k=a;k<=b;k++){
                for(int j=1;j*j<=k;j++){
                    if(k%j==0){
                        S.insert(j);
                        S.insert(k/j);
                    }
                }
            }
            
            for(set<int>::iterator it=S.begin();it!=S.end();it++){
                cnt[*it]++;
            }
            
        }else{
            add(1,b-a+1+1,1,0,0,n);
            add(a,b+1,1,0,0,n);
            int now=2;
            while(1){
                int l=(a+now-1)/now,r=b/now;
                if(l<=b-a+1){
                    if(r>b-a+1) add(b-a+1+1,r+1,1,0,0,n);
                    break;
                }else{
                    if(l!=r) add(l,r+1,1,0,0,n);
                    else cnt[l]++;
                }
                now++;
            }
        }
    }
    
    for(int i=1;i<=M;i++){
        cout<<query(i,i+1,0,0,n)+cnt[i]<<"\n";
    }
}