Description

Cai1000 has $n$ resistors with a resistance of $1\Omega$, and he wants to connect the $n$ resistors as follows:

• Choose a resistor as the original circuit.
• Perform the following operations $n-1$ times: Connect a resistor with a resistance of $1\Omega$ in series or parallel to the previous circuit.

Cai1000 wants to know the $k$-th largest resistance value of the final circuit among all ways of connection (Two ways are considered different if and only if their resistance values are different). Cai1000 is very lazy, so he turned to you for help.

Two resistors with resistances $R_1$ and $R_2$ connected in series equal to a resistor with a resistance of $R=R_1+R_2$.

Two resistors with resistances $R_1$ and $R_2$ connected in parallel equal to a resistor with a resistance of $R=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}$.

Input

One line with two integers $n$ $(1\leq n \leq 60)$ and $k$ $(1\leq k\leq 2^{n-1})$.

Output

Print the $k$-th largest resistance value which is expressed in the form of an irreducible fraction $p/q$, i.e. $p$ and $q$ are integers which meet $p,q\in \mathbb{Z}$ and $\gcd(p,q)=1$.

Samples

3 3

2/3

4 3

5/3

Note

In the first example, there are $4$ ways of connection:

Their resistance values are $\frac{1}{3},\frac{3}{2},\frac{2}{3},3$ respectively. So the third largest resistance value is $\frac{2}{3}$.

Resources

电子科技大学第十四届 ACM 趣味程序设计竞赛