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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from functools import lru_cache
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split())
R = 10**9 + 7
DBG = True  and False
def ddprint(x):
  if DBG:
    print(x)

# extended gcd and modinv
# x = y*(x/y) + x%y
#   ->  g = ax+by = a(y*(x/y) + x%y) + by
#         = (a*(x/y) + b)*y + a*(x%y)
#
# while  g=ax+by, b==0  ->  g=x, a=1

def exgcd(x,y):
    if y == 0:
        return (x,1,0)
    else:
        g,b,a = exgcd(y, x%y)
        return (g, a, b-(x//y)*a)

# return y s.t. x*y mod p == 1
@lru_cache(maxsize=None)
def modinv(x,p):
    g,a,b = exgcd(x,p)  # ax+bp = g, g=1 if mutually prime
    return a%p

@lru_cache(maxsize=None)
def modr(x):
    return x%R

def rmod(i,qb,qw):
    numer = (two2i[i] - qb + qw)
    denom = two2i[i+1]
    iden = modinv(denom, R)
    ddprint("num {} den {} idn {}".format(numer, denom, iden))
    return (numer*iden)%R

b,w = inm()
ddprint("b {} w {}".format(b,w))
two2i = [1] * (b+w+2)
for i in range(1,b+w+2):
    x = (2*two2i[i-1])
    two2i[i] = x if x < R else x-R
cb = cw = 1
qb = qw = 0
print(500000004)
for i in range(1,b+w):
    qb = 0 if i<b else 1 if i==b else (2*qb+cb) % R
    qw = 0 if i<w else 1 if i==w else (2*qw+cw) % R
    cb = 1 if i<b else ((cb*i) * modinv(i-b+1, R)) % R
    cw = 1 if i<w else ((cw*i) * modinv(i-w+1, R)) % R
    ddprint("i {} cb {} cw {} qb {} qw {}".format(i,cb,cw,qb,qw))
    print(rmod(i,qb,qw))