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;

#define NDEBUG
#include <cassert>


typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a)  int((a).size())
#define pb  push_back
#define eb  emplace_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))
#define cons make_pair

#define PQ(T) priority_queue<T>
#define RPQ(T) priority_queue<T,vector<T>,greater<T>>

template<class T> inline void amin(T & a, T const & b) { a = min(a, b); }
template<class T> inline void amax(T & a, T const & b) { a = max(a, b); }

inline ll square(ll x) { return x * x; }
inline ll gcd(ll a, ll b) { while(a) swap(a, b%=a); return b; }
template <typename T>
inline T mod(T a, T b) { return ((a % b) + b) % b; }


const ll MOD=1000000007LL;

ll ADD(ll x, ll y) { return (x+y) % MOD; }
ll SUB(ll x, ll y) { return (x-y+MOD) % MOD; }
ll MUL(ll x, ll y) { return x*y % MOD; }
ll POW(ll x, ll e) { ll v=1; for(; e; x=MUL(x,x), e>>=1) if (e&1) v = MUL(v,x); return v; }
ll INV(ll x) { return POW(x, MOD-2); }
ll DIV(ll x, ll y) { /*assert(y%MOD!=0);*/ return MUL(x, INV(y)); }
ll f[200001];

ll nCr(int n, int r){
    return DIV(f[n], MUL(f[n-r], f[r]));
}

ll solve(int N, ll A, ll B, ll C) {
    ll shrink = DIV(A+B+C, A+B);

    A = DIV(A, 100);
    B = DIV(B, 100);
    C = DIV(C, 100);

    A = MUL(A, shrink);
    B = MUL(B, shrink);

    ll ans = 0;
    {
        ll p = POW(A, N-1), q = p;
        rep(u,N) {
            ll r = MUL(nCr(N-1+u, u), MUL(q, A));
            (ans += MUL(r, N+u)) %= MOD;
            (q *= B) %= MOD;
        }
    }
    {
        ll p = POW(B, N-1), q = p;
        rep(u,N) {
            ll r = MUL(nCr(N-1+u, u), MUL(q, B));
            (ans += MUL(r, N+u)) %= MOD;
            (q *= A) %= MOD;
        }
    }

    ans = MUL(ans, shrink);
    return ans;
}

int main() {
    f[0]=f[1]=1;
    for(int i=2; i<=200000; ++i) {
        f[i] = (f[i-1] * i)% MOD;
    }

    int N,A,B,C; cin >> N >> A >> B >> C;
    cout << solve(N,A,B,C) << endl;
    return 0;
}