Shifty Sum

Take a number such as 12. Shifting a number means appending one zero to its end. Shifting 12 once gives 120, and shifting the result again gives 1200. You may shift a number as many times as you want.

In this problem you compute a shifty sum: the sum of a number and of the numbers obtained by shifting it. You are given a starting number $N$ and a non-negative integer $k$. Add together $N$ and every number obtained by shifting $N$ once, twice, and so on up to $k$ times, that is $N + 10N + 100N + \dots + 10^k N$.

For example, the shifty sum of $N = 12$ with $k = 1$ is $12 + 120 = 132$.

The first line contains the number $N$. ($1 \le N \le 10000$)

The second line contains $k$, the number of times to shift $N$. ($0 \le k \le 5$)

Print one integer, the shifty sum of $N$ with $k$ shifts.