CF868D.Huge Strings

You are given n strings _s_1, s_2, ..., s__n consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation s__a__i__s__b__i is saved into a new string s__n + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2_k such strings) are substrings of the new string. If there is no such k, print 0.

给你 $n$ 个仅由字符 0 和 1 组成的字符串 $s_1, s_2, \dots, s_n$。接下来执行 $m$ 次操作，每次操作将两个已存在的字符串拼接成一个新字符串。在第 $i$ 次操作中，将字符串 $s_{a_i}s_{b_i}$（即 $s_{a_i}$ 与 $s_{b_i}$ 的拼接）保存为新字符串 $s_{n+i}$（操作编号从 1 开始）。每次操作后，你需要找出最大的正整数 $k$，使得所有长度为 $k$ 的、由 0 和 1 构成的字符串（共 $2^k$ 个）均为该新字符串的子串。若不存在这样的 $k$，则输出 0。

The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings _s_1, _s_2, ..., s__n (1 ≤ |s__i| ≤ 100), one per line. The total length of strings is not greater than 100.

The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers a__i abd b__i (1 ≤ a__i, b__i ≤ n + i - 1) — the number of strings that are concatenated to form s__n + i.

第一行包含一个整数 $n$（$1 \leq n \leq 100$）—— 字符串的数量。接下来的 $n$ 行每行包含一个字符串 $s_1,\ s_2,\ \dots,\ s_n$（$1 \leq |s_i| \leq 100$）。所有字符串的总长度不超过 $100$。

下一行包含一个整数 $m$（$1 \leq m \leq 100$）—— 操作的数量。随后有 $m$ 行，每行包含两个整数 $a_i$ 和 $b_i$（$1 \leq a_i,\ b_i \leq n + i - 1$），表示用于构成 $s_{n+i}$ 的两个字符串的编号（即 $s_{n+i} = s_{a_i} + s_{b_i}$）。

Print m lines, each should contain one integer — the answer to the question after the corresponding operation.

输出 m 行，每行应包含一个整数——对应操作后问题的答案。

On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.

On the second operation the string "01100" is created. Now all strings of length k = 2 are present.

On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.

第一次操作创建了一个新字符串“0110”。当 $k = 1$ 时，所有长度为 $k$ 的二进制字符串（即“0”和“1”）均为此新字符串的子串；而当 $k = 2$ 及更大时，存在某些长度为 $k$ 的字符串未在此字符串中出现（例如当 $k = 2$ 时，“00”即为一个未出现的字符串）。因此答案为 1。

第二次操作创建了字符串“01100”。此时所有长度为 $k = 2$ 的二进制字符串均已出现。

第三次操作创建了字符串“1111111111”。该字符串中不含字符“0”，因此答案为 0。