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;
template<class T,class U> inline bool chmin(T&x,U y){if(x>y){x=y;return true;}return false;}
template<class T,class U> inline bool chmax(T&x,U y){if(x<y){x=y;return true;}return false;}
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

struct BIT{
    vector<ll> bt;
    size_t n,m;
    explicit BIT(size_t n):n(n),m(1){
        while(m<=n) m<<=1u;
        m>>=1u;
        bt.resize(n+1);
    }
    ll sum(ll i){//sum of [1,i](1-index)
        if(!i) return 0;
        ll tmp{};
        for(size_t k=i;k;k-=k&-k) tmp+=bt[k];
        return tmp;
    }
    void ad(ll i,ll x){//add to i(1-index)
        for(size_t k=i;k<=n;k+=k&-k) bt[k]+=x;
    }
    ll lb(ll k){//if sum(i)<k and sum(i) >= k, return i
        ll x=0;
        for(size_t i=m;i>0;i>>=1u){
            if(x+i<=n&&bt[x+i]<k) {
                k-=bt[x+i];
                x+=i;
            }
        }
        return x+1;
    }
};

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    vector<pair<int,int>> p(n);
    for(auto&lr:p) cin>>lr.first>>lr.second;
    sort(p.begin(),p.end(),[](auto x,auto y){
        return x.second-x.first<y.second-y.first;
    });
    BIT bt(m+1);
    int s{},t{n},tmp;
    Fr(d,m){
        while(s<n&&p[s].second-p[s].first+1<d){
            --t;
            bt.ad(p[s].first,1);
            bt.ad(p[s].second+1,-1);
            ++s;
        }
        tmp=0;
        for(int i=d;i<=m;i+=d) tmp+=bt.sum(i);
        cout<<t+tmp<<endl;
    }
}