Old teletext

In this problem, we use a simplified model of teletext, where the
available pages are 100–899, the initial page is always 100, and there
are only two ways to move to any page:

- By pressing its three digits.

- By (repeatedly) pressing the ‘+’ key or the ‘-’ key. The ‘+’ key
  increments the current page number by one (or from 899 to 100). The
  ‘-’ key decrements the current page number by one (or from 100 to
  899).

Professor Oak’s television is so old, that some keys of its remote
control are broken. For instance, assume that the keys ‘1’ and ‘-’ are
broken. If he wants to change from page 140 to page 211, he cannot press
‘211’ directly. Still, he can just press ‘+’ 71 times. Alternatively, he
can press ‘200’, and afterwards press ‘+’ 11 times, for a total cost of
14. For this example, the minimum cost is 5: first press ‘209’; then
press ‘+’ twice.

Given a list of broken keys, and a list of teletext pages that must be
visited, which is the minimum cost to visit all those pages at least
once, in any order?

Input

Input consists of at most 100 cases. Every case has two lines. The first
line has a string with all the broken keys in any order, following
exactly the sample input. The second line has a number p, followed by p
page numbers with no repetitions. Assume 1 ≤ p ≤ 9.

Output

For every case, print the minimum cost. Print “no solution” if there is
none.

Problem information

Author: Salvador Roura

Generation: 2026-01-25T11:57:25.233Z

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