Football rivalry (2) P94654

Two long-time rival football teams, let us call them $B$ (for beautiful manners) and $M$ (for miserable — very, very miserable — manners), are playing again. Both teams are exhausted, so the first to score a goal will win the game for sure. At this moment, team $B$ has the ball. If they decide to attack, there is a probability $w_B$ that they manage to score, thus winning the game. Hovewer, with probability $\ell_B$ they will receive a goal, thus losing the game. With probability $1 - w_B - \ell_B$ they will just lose the possesion of the ball. Team $B$ has another option: to pass the ball around. In that case, the possesion of the ball will eventually go to team $M$. Then we will have a simmetrical situation: If team $M$ goes for an attack, they will immediately win with probability $w_M$, they will immediately lose with probability $\ell_M$, and the ball will go back to team $B$ with probability $1 - w_M - \ell_M$. If they decide to just pass the ball and wait, eventually the possesion of the ball will go back to team $B$.

0.55

Given $w_B$, $\ell_B$, $w_M$ and $\ell_M$, and assuming that both teams take the best decisions (to attack or not to attack) and that team $B$ has the ball now, which is the probability that team $B$ will win?

Input

Input consists of several cases, each one with four real numbers $w_B$, $\ell_B$, $w_M$ and $\ell_M$ between 0 and 1. Assume $w_B + \ell_B \le 1$ and $w_M + \ell_M \le 1$.

  0.45

Output

For every case, print the probability that team $B$ will win with four digits after the decimal point. (The input cases have no precision issues.) A situation where no goal will be scored (an eternal tie) is similar to a fifty-fifty situation. Consequently, print “0.5000” in this case.