Problem:
Score : 500 points

Problem Statement

Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total.

When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %.
Find the expected number of games that will be played, and print it as follows.

We can represent the expected value as P/Q with coprime integers P and Q.
Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}.
(Such an integer R always uniquely exists under the constraints of this problem.)

Constraints

1 \leq N \leq 100000

0 \leq A,B,C \leq 100

1 \leq A+B

A+B+C=100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C

Output

Print the expected number of games that will be played, in the manner specified in the statement.

Sample Input 1

1 25 25 50

Sample Output 1

2

Since N=1, they will repeat the game until one of them wins.
The expected number of games played is 2.

Sample Input 2

4 50 50 0

Sample Output 2

312500008

C may be 0.

Sample Input 3

1 100 0 0

Sample Output 3

1

B may also be 0.

Sample Input 4

100000 31 41 28

Sample Output 4

104136146

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

// test

import java.util.*;
class Main {
    public static final int mod = 1_000_000_007;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();
        int c = sc.nextInt();
        long ans = 0;

        long aPct = div(a, a+b);
        long bPct = div(b, a+b);
        long comb = 1;
        for(int m = 0; m<n; m++) {
            ans += pow(aPct, n)*pow(bPct, m)%mod*comb%mod*(n+m)%mod;
            ans %= mod;
            ans += pow(bPct, n)*pow(aPct, m)%mod*comb%mod*(n+m)%mod;
            ans %= mod;
            comb = div(comb*(n+m)%mod, m+1);
        }
        long l = div(a+b, 100);
        System.out.println(div(ans, l));

    }
    static long pow (long x, long y) {
        if (y == 0) {
            return 1;
        } else if (y % 2 == 0) {
            long a = pow(x, y/2);
            return a*a%mod;
        } else {
            long a = pow(x, y/2);
                return(a*a%mod)*x%mod;
        }
    }

    static long div (long x, long y) {
        return (x*pow(y, mod-2))%mod;
    }

}