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: Accepted)

MOD = 10 ** 9 + 7
fac = [1] * (10 ** 5 * 3)


def nCr(n, r):
    return fac[n] * pow(fac[n - r], MOD - 2, MOD) * pow(fac[r], MOD - 2, MOD) % MOD


for i in range(len(fac) - 1):
    fac[i + 1] = fac[i] * (i + 1) % MOD
b, w = map(int, input().split())
n = b + w
p = [0] * n
q = [0] * n
for i in range(b, n):
    p[i] = (p[i - 1] + nCr(i - 1, b - 1) *
            pow(pow(2, i, MOD), MOD - 2, MOD)) % MOD
for i in range(w, n):
    q[i] = (q[i - 1] + nCr(i - 1, w - 1) *
            pow(pow(2, i, MOD), MOD - 2, MOD)) % MOD
t = pow(2, MOD - 2, MOD)
for i in range(n):
    print((q[i] + (1 - p[i] - q[i]) * t) % MOD)