Positional number systems

We are used to writing numbers in decimal. Writing "123" really means $1 \cdot 10^2 + 2 \cdot 10 + 3$.

Binary shows up often too. The binary form of $123$ is "1111011", which means $1 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2 + 1$.

The base of a positional system does not have to be a natural number. Written in base $-10$, the number $123$ becomes "283", which means $2 \cdot (-10)^2 + 8 \cdot (-10) + 3$.

The base does not even have to be an integer. In base $2.5$ the number $123$ is written "22122.02012122$\ldots$", where the part after the point runs on forever to the right.

Bases smaller than $1$ make sense as well. In such a system the notation comes out mirrored, because place values grow as you move right from the point. In base $0.1$ the number $123$ is written "3.21", which means $3 + 2 \cdot 0.1^{-1} + 1 \cdot 0.1^{-2}$.

You are given an integer $R$ and a base $B$. Print the representation of $R$ in base $B$.

The first line has the integer $R$ in decimal ($0 \le R \le 10^{10} - 1$), at most $10$ digits.

The second line has the base $B$ in decimal. Either $B$ is an integer with $2 \le |B| \le 10$, or $1/B$ is an integer with $2 \le |1/B| \le 10$.

Let $b = B$ when $|B| > 1$, and $b = 1/B$ when $|B| < 1$. In both cases $b$ is an integer, and the representation uses the digits $0$ to $|b| - 1$.

When $|B| > 1$, print the digits of $R$ in base $B$ on one line, from the highest place value down to the $B^0$ place. Print no point and no leading zero.

When $|B| < 1$, print the digit of the $B^0$ place first, then a point, then the digits of the $B^{-1}$, $B^{-2}$, $\ldots$ places. The last digit printed is not $0$. If there is only one digit, print it without a point.

If $R = 0$, print "0".

Exactly one representation follows these rules, and it equals $R$ with no rounding.