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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

def comb(n,k,p):
  """power_funcを用いて(nCk) mod p を求める"""
  from math import factorial
  if n<0 or k<0 or n<k: return 0
  if n==0 or k==0: return 1
  a=factorial(n) %p
  b=factorial(k) %p
  c=factorial(n-k) %p
  return (a*power_func(b,p-2,p)*power_func(c,p-2,p))%p
 
def power_func(a,b,p):
  """a^b mod p を求める"""
  if b==0: return 1
  if b%2==0:
    d=power_func(a,b//2,p)
    return d*d %p
  if b%2==1:
    return (a*power_func(a,b-1,p ))%p
B, W = map(int,input().split())
Mod = 10**9+7
magick = 500000004

consts_B, consts_W = [], []
for i in range(B+W):
  consts_B += [comb(i,B-1,Mod)]
  consts_W += [comb(i,W-1,Mod)]

p, q = 0, 0
inv = 1
for i in range(B+W):
  print((1-p+q)*magick%Mod)
  inv *= magick
  p += consts_B[i]*inv
  q += consts_W[i]*inv