output_description: Print the expected number of games that will be played, in the manner
  specified in the statement.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N A B C
attempt: # 入力
  N, A, B, C = map(int, input().split())
  
  MOD = 10**9 + 7
  
  
  class ModInt:
      def __init__(self, x):
          self.x = x % MOD
  
      def __str__(self):
          return str(self.x)
  
      __repr__ = __str__
  
      def __add__(self, other):
          return (
              ModInt(self.x + other.x)
              if isinstance(other, ModInt)
              else ModInt(self.x + other)
          )
  
      def __sub__(self, other):
          return (
              ModInt(self.x - other.x)
              if isinstance(other, ModInt)
              else ModInt(self.x - other)
          )
  
      def __mul__(self, other):
          return (
              ModInt(self.x * other.x)
              if isinstance(other, ModInt)
              else ModInt(self.x * other)
          )
  
      def __truediv__(self, other):
          return (
              ModInt(self.x * pow(other.x, MOD - 2, MOD))
              if isinstance(other, ModInt)
              else ModInt(self.x * pow(other, MOD - 2, MOD))
          )
  
      def __pow__(self, other):
          return (
              ModInt(pow(self.x, other.x, MOD))
              if isinstance(other, ModInt)
              else ModInt(pow(self.x, other, MOD))
          )
  
      def __radd__(self, other):
          return ModInt(other + self.x)
  
      def __rsub__(self, other):
          return ModInt(other - self.x)
  
      def __rmul__(self, other):
          return ModInt(other * self.x)
  
      def __rtruediv__(self, other):
          return ModInt(other * pow(self.x, MOD - 2, MOD))
  
      def __rpow__(self, other):
          return ModInt(pow(other, self.x, MOD))
  
  
  # f[n] = n! % (10**9 + 7)
  f = [0 for _ in range(2 * N)]
  f[0] = ModInt(1)
  for i in range(1, 2 * N):
      f[i] = i * f[i - 1]
  
  # 引き分けなしの分母
  D = ModInt(100 - C)
  # 1回のゲームにつき、高橋君が勝つ確率
  p = A / D
  # 1回のゲームにつき、青木君が勝つ確率
  q = B / D
  
  # 高橋君がN回、青木君がi回勝つケースの重み付き期待値の和、および
  # 高橋君がi回、青木君がN回勝つケースの重み付き期待値の和
  # を求めそれらの和を解とする
  ans = (100 / D) * sum(
      (N + i) * (f[N + i - 1] / (f[i] * f[N - 1])) * (p**N * q**i + p**i * q**N)
      for i in range(N)
  )
  
  # 出力
  print(ans)
  
problem_description: 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.)
samples: [{"input": "1 25 25 50", "output": "2\n    \n\nSince N=1, they will repeat the game until one of them wins. The expected\nnumber of games played is 2.\n\n* * *"}, {"input": "4 50 50 0", "output": "312500008\n    \n\nC may be 0.\n\n* * *"}, {"input": "1 100 0 0", "output": "1\n    \n\nB may also be 0.\n\n* * *"}, {"input": "100000 31 41 28", "output": "104136146"}]