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.

Examples

Input

1 25 25 50


Output

2


Input

4 50 50 0


Output

312500008


Input

1 100 0 0


Output

1


Input

100000 31 41 28


Output

104136146
#include <bits/stdc++.h>

typedef long long int ll;
typedef unsigned long long int ull;
#define rep(i,n) for(auto i=0;i<(n);i++)
#define rep1(i,n) for(auto i=1;i<=(n);i++)
#define all(x) (x).begin(),(x).end()

using namespace std;


struct init{ init(){ cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(20); }; }aaaaaaa;


ll mod = 1e9 + 7;

ll powmod(ll b){
  ll r = 1;
  ll c = mod-2;
  for (; c>0; c>>=1, b=b*b%mod){
    if (c&1)  r = r*b%mod;
  }
  return r;
}

ll powdiv(ll a, ll b){
  return a * powmod(b) % mod;
}

int main(){
  ll n, a, b, c;
  cin >> n >> a >> b >> c;

  ll ab = a+b;
  a += powdiv(c * a, ab);
  b += powdiv(c * b, ab);

  ll comp = 1;
  ll p = 0;
  
  ll t = 1;
  rep(i, n-1) t *= 100, t %= mod;
  
  ll an = 1;
  rep(i, n) an *= a, an %= mod;
  ll bn = 1;
  rep(i, n) bn *= b, bn %= mod;

  ll ai = 1;
  ll bi = 1;


  rep(i, n){
    if(0 < i){
      comp *= n-1+i, comp %= mod;
      comp = powdiv(comp, i), comp %= mod;
    }
    ll s = ((((n+i) * t)%mod) * comp)%mod;
    p += (((an * bi)%mod) * s)%mod;
    p %= mod;
    p += (((bn * ai)%mod) * s)%mod;
    p %= mod;
    ai *= a, ai %= mod;
    bi *= b, bi %= mod;
    t = powdiv(t, 100);
  }
  
  ll res = p;
  res = powdiv(res, 100-c);
  rep(i, 2*n-2) res = powdiv(res, 100);

  cout << res << endl;



  return 0;
}
