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: Accepted)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod=1e9+7;
ll B,W,inv[200005],f[200005],p[200005],w[200005],b[200005];
ll C(ll n,ll m)
{
    ll ans=f[n]*inv[n-m]%mod*inv[m]%mod;
    return ans;
}
ll qpow(ll a,ll n)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=ans*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ans;
}
int main()
{
    scanf("%lld%lld",&B,&W);
    f[0]=1;inv[0]=1;
    for(int i=1;i<=B+W;i++)
        f[i]=f[i-1]*i%mod,inv[i]=qpow(f[i],mod-2);
    p[0]=1;
    for(int i=1;i<=B+W;i++)
        p[i]=p[i-1]*(mod-mod/2)%mod;
    for(int i=B;i<B+W;i++)
        b[i]=p[i]*C(i-1,B-1)%mod;
    for(int i=W;i<B+W;i++)
        w[i]=p[i]*C(i-1,W-1)%mod;
    for(int i=1;i<=B+W;i++)
    {
        w[i]=(w[i]+w[i-1])%mod;
        b[i]=(b[i]+b[i-1])%mod;
    }
    for(int i=1;i<=B+W;i++)
    {
        ll ans=w[i-1]+(1-w[i-1]-b[i-1])*(i==B+W?1:(mod-mod/2))%mod;
        printf("%lld\n",(ans%mod+mod)%mod);
    }
}