Alternative Scale of Notation

We can map strings over an alphabet $\Sigma_B = \{C_1, C_2, \dots, C_B\}$ of size $B$ to non-negative integers. Treat the characters as digits $C_1 = 0, C_2 = 1, \dots, C_B = B - 1$ and read the string as a base-$B$ numeral. Call this map $U_B$: for a string $\alpha[1..n]$ of length $n$,

$U_B(\alpha) = \sum_{i=0}^{n-1} \alpha[n-i] \cdot B^i.$

For example, $U_3(1001) = 1 \cdot 27 + 0 \cdot 9 + 0 \cdot 3 + 1 \cdot 1 = 28$.

This correspondence has a serious drawback: it is not one-to-one. For example,

$28 = U_3(1001) = U_3(01001) = U_3(001001) = \dots,$

so infinitely many strings map to $28$.

To avoid this, we use an alternative interpretation. We still read characters as digits, but shifted by one: $C_1 = 1, C_2 = 2, \dots, C_B = B$. There is no digit $0$ anymore; instead there is a digit $B$. We define the map $V_B$ in the same shape: for a string $\alpha[1..n]$ of length $n$,

$V_B(\alpha) = \sum_{i=0}^{n-1} \alpha[n-i] \cdot B^i.$

For the empty string $\epsilon$ we set $V_B(\epsilon) = 0$.

This looks a lot like $U_B$, but the digit set is different. For example, $V_3(1313) = 1 \cdot 27 + 3 \cdot 9 + 1 \cdot 3 + 3 \cdot 1 = 60$.

It can be shown that this map is bijective (one-to-one and onto), so it has an inverse. Your task is to compute the inverse of $V_B$: given an integer $x$, find the string $\alpha$ such that $V_B(\alpha) = x$.

The first line contains $B$ ($2 \le B \le 9$).

The second line contains an integer $x$ written in ordinary decimal notation, with $0 \le x \le 10^{100}$.

Print a string $\alpha$ consisting only of digits from $\{1, 2, \dots, B\}$ such that $V_B(\alpha) = x$. The answer is unique. When $x = 0$, the string is empty, so print nothing.