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

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
ll modpow(ll x,ll n){
	ll res=1;
	while(n>0){
		if(n&1) res=res*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return res;
}
ll F[200005],R[200005];
void make(){
	F[0] = 1;
	for(int i=1;i<200005;i++) F[i] = F[i-1]*i%mod;
	for(int i=0;i<200005;i++) R[i] = modpow(F[i],mod-2);
}
ll C(int a,int b){
	return F[a]*R[b]%mod*R[a-b]%mod;
}
ll n,a,b,c;
int main(){
	cin>>n>>a>>b>>c; make();
	ll ans = 0;
	for(int i=0;i<n;i++){
		ll up = modpow(a,n)*modpow(b,i)%mod*C(n+i-1,i)%mod;
		up = up*modpow(modpow(a+b,n+i),mod-2)%mod; //cout<<up<<endl;
		up = up*(n+i)%mod*100LL%mod*modpow(a+b,mod-2)%mod;
		ans += up;
		up = modpow(b,n)*modpow(a,i)%mod*C(n+i-1,i)%mod;
		up = up*modpow(modpow(a+b,n+i),mod-2)%mod;
		up = up*(n+i)%mod*100LL%mod*modpow(a+b,mod-2)%mod;
		ans += up;
	}
	cout << (ans%mod+mod)%mod << endl;
}