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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>
#include<string.h>
#include<math.h>

typedef int32_t i32;
typedef int64_t i64;

const i32 mod = 1000000007;

i32 inv (i32 a) {
  i32 t = 1;
  while (a > 1) {
    t = (i64) t * (mod - mod / a) % mod;
    a = mod % a;
  }
  return t;
}

i32 modPow(i32 r, i32 n) {
  i32 t = 1;
  i32 s = r;
  while (n > 0) {
    if (n&1) t = (i64) t * s % mod;
    s = (i64) s * s % mod;
    n >>= 1;
  }
  return t;
}

i32 *fact = NULL;
i32 *iFact = NULL;
void init(const i32 n) {
  fact = (i32 *) calloc (n + 1, sizeof (i32));
  fact[0] = 1;
  for (i32 i = 1; i <= n; ++i) {
    fact[i] = (i64) i * fact[i - 1] % mod;
  }
  iFact = (i32 *) calloc (n + 1, sizeof (i32));
  iFact[n] = inv(fact[n]);
  for (i32 i = n - 1; i >= 0; --i) {
    iFact[i] = (i64) iFact[i + 1] * (i + 1) % mod;
  }
}

i32 comb (i32 n, i32 k) {
  if (!(0 <= k && k <= n)) return 0;
  return (i64) fact[n] * iFact[k] % mod * iFact[n - k] % mod;
}

void run (void) {
  i32 b, w;
  scanf ("%" SCNi32 "%" SCNi32, &b, &w);
  init(b + w);
  i32 l = 0;
  i32 r = 0;
  const i32 p = inv(2);
  for (i32 i = 1; i <= b + w; ++i) {
    printf ("%" PRIi64 "\n", (((i64)2 * mod + 1 - r - l) * p + l) % mod);
    if (i >= w) {
      l = (l + (i64)comb (i - 1, w - 1) * modPow (p, i)) % mod;
    }
    if (i >= b) {
      r = (r + (i64)comb (i - 1, b - 1) * modPow (p, i)) % mod;
    }
  }
}

int main (void) {
  run ();
  return 0;
}