Problem:
Score : 700 points

Problem Statement

Today, Snuke will eat B pieces of black chocolate and W pieces of white chocolate for an afternoon snack.

He will repeat the following procedure until there is no piece left:

Choose black or white with equal probability, and eat a piece of that color if it exists.

For each integer i from 1 to B+W (inclusive), find the probability that the color of the i-th piece to be eaten is black.
It can be shown that these probabilities are rational, and we ask you to print them modulo 10^9 + 7, as described in Notes.

Notes

When you print a rational number, first write it as a fraction \frac{y}{x}, where x, y are integers and x is not divisible by 10^9 + 7
(under the constraints of the problem, such representation is always possible).
Then, you need to print the only integer z between 0 and 10^9 + 6, inclusive, that satisfies xz \equiv y \pmod{10^9 + 7}.

Constraints

All values in input are integers.

1 \leq B,W \leq 10^{5}

Input

Input is given from Standard Input in the following format:

B W

Output

Print the answers in B+W lines. In the i-th line, print the probability that the color of the i-th piece to be eaten is black, modulo 10^{9}+7.

Sample Input 1

2 1

Sample Output 1

500000004
750000006
750000006

There are three possible orders in which Snuke eats the pieces:

white, black, black

black, white, black

black, black, white

with probabilities \frac{1}{2}, \frac{1}{4}, \frac{1}{4}, respectively. Thus, the probabilities of eating a black piece first, second and third are \frac{1}{2},\frac{3}{4} and \frac{3}{4}, respectively.

Sample Input 2

3 2

Sample Output 2

500000004
500000004
625000005
187500002
187500002

They are \frac{1}{2},\frac{1}{2},\frac{5}{8},\frac{11}{16} and \frac{11}{16}, respectively.

Sample Input 3

6 9

Sample Output 3

500000004
500000004
500000004
500000004
500000004
500000004
929687507
218750002
224609377
303710940
633300786
694091802
172485353
411682132
411682132

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>  
#include <algorithm>  
#include <assert.h>
#include <bitset>
#include <cmath>  
#include <complex>  
#include <deque>  
#include <functional>  
#include <iostream>  
#include <limits.h>  
#include <map>  
#include <math.h>  
#include <queue>  
#include <set>  
#include <stdlib.h>  
#include <string.h>  
#include <string>  
#include <time.h>  
#include <unordered_map>  
#include <unordered_set>  
#include <vector>  
#include <chrono>
#include <random>
#include <time.h>
#define ll long long
#define rep2(i,a,b) for(ll i=a;i<=b;++i)
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep3(i,a,b) for(ll i=a;i>=b;i--)
#define REP(e,v) for(auto e:v)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define tii tuple<int,int,int>
#define pq priority_queue<int>
#define pqg priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define edge(v,a,b) v[a].pb(b);v[b].pb(a);
#define MAX_V 400010
#define vec vector<int>
#define vecll vector<ll>
#define vecpii vector<pii>
#define endl "\n"
#define ALL(c) (c).begin(),(c).end()
using namespace std;
int in() {int x;scanf("%d",&x);return x;}
ll lin() {ll x;scanf("%lld",&x);return x;}

#define INF 1e9+7
#define LLINF 1e18+7
ll MOD=1e9+7;
#define N 210000
ll int inv[N],comb[N],prd[N],invprd[N];
void calc_inv(){
    inv[0]=1;
    inv[1]=1;
    rep2(i,2,N){
        inv[i]=((-(MOD/i)*inv[MOD%i])%MOD+MOD)%MOD;
    }
    return;
}
void calc_product(){
    prd[0]=prd[1]=1;
    invprd[0]=invprd[1]=1;
    rep2(i,2,N-1){
        prd[i]=i*prd[i-1]%MOD;
        invprd[i]=inv[i]*invprd[i-1]%MOD;
    }
    return ;
}
ll cmb(int a,int b){
    return prd[a]*invprd[b]%MOD*invprd[a-b]%MOD;
}
ll inv2[N];
main(){
    calc_inv();
    calc_product();
    inv2[0]=1;
    rep(i,N-1){
        inv2[i+1]=inv2[i]*inv[2]%MOD;
    }
    ll ans[N];
    int b=in(),w=in();
    rep2(i,1,b+w)ans[i]=inv[2];
    ll now=0;
    rep2(i,w+1,b+w){
        now=(now+inv2[i]*cmb(i-2,w-1)%MOD)%MOD;
        cout<<cmb(i-1,w)<<endl;
        ans[i]=(ans[i]+now)%MOD;
    }
    now=0;
    rep2(i,b+1,b+w){
        now=(now+inv2[i]*cmb(i-2,b-1)%MOD)%MOD;
        ans[i]=(ans[i]+MOD-now)%MOD;
    }
    rep2(i,1,b+w){
        cout<<ans[i]<<endl;
    }
}