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 <iostream>
#include <cassert>
#include <climits>
#include <bitset>
#include <stack>
#include <queue>
#include <iomanip>
#include <limits>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <tuple>
#include <time.h>

using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;

long long int INF = 3e18;
double Pi = 3.1415926535897932384626;

vector<ll> G[500005];
vector<P> tree[500010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop

int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//ll bit[500005];
//↓,→,↑,←


#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<(x)<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pd(x) cerr<<"//"<<(x)<<endl;
#define re(i,n) for(i=0;i<n;i++);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define deba(x) cout<< #x << " = " << x <<endl


ll mod = 1000000007;
ll rui(ll number1,ll number2){
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){
    if(number1 > number2){
        swap(number1,number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1,number1);
    }
}
void YES(bool condition){
    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;
}
void Yes(bool condition){
    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;
}

ll n,m,num,sum,ans,a,b,c,d,e,g,h,w,i,j,k,l,q;
ll x[500005],y[500005],z[500005];
ll dp[5][200005];
char s[500005],t[500005];
bool flag,dame;

ll  fact[400005];
ll rfact[400005];

void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<400000; i++){
        fact[i] = (fact[i-1]*i) % mod;
        rfact[i] = rui(fact[i],mod - 2);
    }
}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod) * rfact[n-r]) % mod;
}


int main(){
    cin >> b >> w;
	c3_init();
	for(i=0;i<b+w;i++){
		if(i == 0){
			dp[1][i] = 1;
			y[i] = 500000004ll;
			p(500000004);
		}else{
			if(i < min(b,w)){
				dp[1][i] = dp[1][i-1] * 2;
			}else if(i == b && i < w){
				dp[1][i] = dp[1][i-1] * 2 + mod - 1;
				dp[0][i] = 1;
			}else if(i == w && i < b){
				dp[1][i] = dp[1][i-1] * 2 + mod - 1;
				dp[2][i] = 1;
			}else if(i == b && i == w){
				dp[1][i] = dp[1][i] * 2 + mod - 2;
				dp[0][i] = dp[2][i] = 1;
			}else{
				//dp[0][i] = dp[0][i-1] * 2 + c3(i-1,i-b);
				//dp[2][i] = dp[2][i-1] * 2 + c3(i-1,i-1-(i-w));
				dp[1][i] = (dp[0][i-1] + dp[1][i-1] + dp[2][i-1]) * 2 + (mod - dp[0][i]) + (mod - dp[2][i]);
			}
			dp[0][i] %= mod;
			dp[1][i] %= mod;
			dp[2][i] %= mod;
			//pe(i);
			//p(dp[1][i] + 2 * dp[2][i]);
			y[i] = y[i-1] * 500000004ll;
			y[i] %= mod;
			ans = (((dp[1][i] + 2 * dp[2][i]) % mod) * y[i])%mod;
			p(ans);
		}
	}


	return 0;
}