Problem F
Hackenbush Simplicity

When calculating the values of Hackenbush positions a particular operation comes up quite a lot. It takes two dyadic fractions $x, y$ and produces another dyadic fraction. If there exists an integer $z$ such that $x < z < y$ then the operation returns an integer, the integer $z$ satisfying $x < z < y$ that has the least absolute value. If no such integer exists the operation returns the dyadic fraction $p/2^q$ satisfying $x < p/2^q < y$ having the lowest power $q$ in the denominator. This fraction is uniquely determined.

All fractions in this problem are given in the format x/2q where $x, q$ are integers, additionally $q \geq 0$. If the fraction is an integer /20 is omitted. These fractions are always fully reduced.

Input

The input is two lines, each containing a fraction in the format described above. These fractions are the values $x, y$ which are the inputs of the operation described above. You may assume $x < y$. The exponent in the denominator will be at most $10^9$ and the numerator will have at most $10^4$ digits.

Output

Print the output of applying the operation to $x, y$. Print it in the format described above.

3/2^2
7/2^3

13/2^4

-1/2^2
1/2^9

0

9/2^2
19/2^2

3

1
2

3/2^1