K-equivalence

Consider a set $K$ of positive integers.

Let $p$ and $q$ be two non-zero decimal digits. We call them $K$-equivalent when the following holds:

For every $n \in K$, replacing a single digit $p$ by $q$, or a single digit $q$ by $p$, in the decimal representation of $n$ always yields a number that is again an element of $K$.

For example, if $K$ is the set of integers divisible by $3$, then the digits $1$, $4$, and $7$ are $K$-equivalent: replacing a $1$ by a $4$ in the decimal representation of a number never changes its divisibility by $3$.

$K$-equivalence is an equivalence relation on the digits (it is reflexive, symmetric, and transitive).

You are given a finite set $K$ expressed as a union of pairwise-disjoint finite intervals of positive integers. Find the equivalence classes of the digits $1$ through $9$.

The first line contains $n$, the number of intervals whose union forms $K$ ($1 \le n \le 10000$).

Each of the next $n$ lines contains two positive integers $a_i$ and $b_i$ that describe the interval $[a_i, b_i]$ (all integers $x$ with $a_i \le x \le b_i$), where $1 \le a_i \le b_i \le 10^{18}$. Moreover, for every $i$ with $2 \le i \le n$, $a_i \ge b_{i-1} + 2$ (so the intervals are pairwise disjoint and given in increasing order).

Represent each equivalence class as the concatenation of its digits in ascending order.

Print all equivalence classes of the digits $1$ through $9$, one per line, sorted lexicographically.