# Original solution
from functools import lru_cache
import sys

sys.setrecursionlimit(2*10**5)

# 入力
B, W = list(map(int, input().split()))

# ModInt定義
MOD = 10**9 + 7


class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    def __repr__(self):
        return str(self.x)

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    __radd__ = __add__
    __rsub__ = __sub__
    __rmul__ = __mul__
    __rtruediv__ = __truediv__
    __rpow__ = __pow__


# 最初にm個あったチョコレートについて、i番目までにすべて食べきる確率と確率の計算に必要な二項係数を返す
@lru_cache(maxsize=None)
def p(i, m):
    if i < m:
        return ModInt(0), ModInt(0)
    elif i == m:
        return ModInt(1) / ModInt(pow(2, i, MOD)), ModInt(1)
    else:
        x, y = p(i - 1, m)
        c = y * (i - 1) / (i - m)
        return x + c / pow(2, i, MOD), c


# i番目の解は i-1番目までに白チョコレートが枯渇している確率 +
# i-1番目までにどちらのチョコレートも枯渇しない確率 / 2
ans = '\n'.join(
    str((1 + p(i - 1, W)[0] - p(i - 1, B)[0]) / 2)
    for i in range(1, B + W + 1)
)

# 出力
print(ans)

# Optimized solution
# 入力
B, W = list(map(int, input().split()))

# ModInt定義
MOD = 10**9 + 7


class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    def __repr__(self):
        return str(self.x)

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    __radd__ = __add__
    __rsub__ = __sub__
    __rmul__ = __mul__
    __rtruediv__ = __truediv__
    __rpow__ = __pow__


p = ModInt(0)
q = ModInt(0)
cp = ModInt(0)
cq = ModInt(0)
for i in range(1, B + W + 1):
    # i番目の解は i-1番目までに白チョコレートが枯渇している確率 +
    # i-1番目までにどちらのチョコレートも枯渇しない確率 / 2
    ans = (1 + q - p) / 2
    # 出力
    print(ans)
    cp = (
        ModInt(0) if i < B else
        ModInt(1) if i == B else
        cp * (i - 1) / (i - B)
    )
    p = (
        ModInt(0) if i < B else
        ModInt(1) / ModInt(2**i) if i == B else
        p + cp / pow(2, i, MOD)
    )
    cq = (
        ModInt(0) if i < W else
        ModInt(1) if i == W else
        cq * (i - 1) / (i - W)
    )
    q = (
        ModInt(0) if i < W else
        ModInt(1) / ModInt(2**i) if i == W else
        q + cq / pow(2, i, MOD)
    )