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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define mod 1000000007
#define MAX_N 200010
long long inv[MAX_N];
long long factorial[MAX_N];
long long inv_factorial[MAX_N];

void GetInv(){
    for (int i = 1; i < MAX_N; i++) {
        if (i == 1) inv[i] = 1;
        else {
            inv[i] = (mod - (mod / i) * inv[mod % i]) % mod;
            if (inv[i] < 0) inv[i] += mod;
        }
    }
}
void GetFactorial(){
    factorial[0] = 1; inv_factorial[0] = 1;
    for (int i = 1; i < MAX_N; i++) {
        factorial[i] = factorial[i-1] * i;
        factorial[i] %= mod;
        inv_factorial[i] = inv_factorial[i-1] * inv[i];
        inv_factorial[i] %= mod;
    }
}

long long combination(int n, int r) {
    long long ret = factorial[n] * inv_factorial[r];
    ret %= mod;
    ret *= inv_factorial[n-r];
    return ret % mod;
}

long long modpow(int n, int r) {
    long long ret = 1; long long tmp = (long long) n;
    while (r != 0) {
        if (r % 2) ret *= tmp;
        tmp *= tmp; tmp %= mod; ret %= mod;
        r /= 2;
    }
    return ret;
}

int main() {
    GetInv(); GetFactorial();
    int b, w; cin >> b >> w;
    int n = b + w;
    vector<long long> p(n+1, 0);
    for (int i = 0; i < b; i++) {
        long long pp = modpow(inv[2], (n-i-1)) * combination(n-i-2, w-1) % mod;
        long long c = pp * (b-i-1) % mod * inv[n-i-2] % mod;
        p[0] = (p[0] + c) % mod;
        p[n-i-2] = (mod + p[n-i-2] - c) % mod;
        p[n-i-1] = (p[n-i-1] + pp) % mod;
        p[n] = (mod + p[n] - pp) % mod; 
    }
    for (int i = 0; i < w; i++) {
        long long pp = modpow(inv[2], (n-i-1)) * combination(n-i-2, b-1) % mod;
        long long c = pp * (b-1) % mod * inv[n-i-2] % mod;
        p[0] = (p[0] + c) % mod;
        p[n-i-2] = (mod + p[n-i-2] - c + pp) % mod;
        p[n-i-1] = (mod + p[n-i-1] - pp) % mod;
    }
    for (int i = 0; i < n; i++) {
        p[i+1] = (p[i+1] + p[i]) % mod;
        cout << p[i] << endl;
    }
    return 0; 
}