Codeforces Round #244 (Div. 2)D（字符串DP）_html/css_WEB-ITnose

D. Match & Catch

time limit per test

memory limit per test

input

output

Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings s1 and s2 from two different frequencies as signals. They are suspecting that these two strings are from two different criminals and they are planning to do some evil task.

Now they are trying to find a common substring of minimum length between these two strings. The substring must occur only once in the first string, and also it must occur only once in the second string.

Given two strings s1 and s2 consist of lowercase Latin letters, find the smallest (by length) common substring p of both s1 and s2, where pis a unique substring in s1 and also in s2. See notes for formal definition of substring and uniqueness.

Input

The first line of input contains s1 and the second line contains s2 (1?≤?|s1|,?|s2|?≤?5000). Both strings consist of lowercase Latin letters.

Output

Print the length of the smallest common unique substring of s1 and s2. If there are no common unique substrings of s1 and s2 print -1.

Sample test(s)

input

applepepperoni

output

input

loverdriver

output

input

bidhanroy

output

-1

input

testsetsesteeptes

output

Note

Imagine we have string a?=?a1a2a3...a|a|, where |a| is the length of string a, and ai is the ith letter of the string.

We will call string alal?+?1al?+?2...ar (1?≤?l?≤?r?≤?|a|) the substring [l,?r] of the string a.

The substring [l,?r] is unique in a if and only if there is no pair l1,?r1 such that l1?≠?l and the substring [l1,?r1] is equal to the substring[l,?r] in a.

题意：RT

思路：用dp1[i][j]表示s1中以i为起始,s2中以j为起始的最长匹配子串

           dp2[i][j]表示s1中以i为起始，s1中以j为起始的最长匹配串

           然后枚举s1的每个位置i，在s2中找出所有与i位置匹配的串

           在s1中找出所有（除了i位置）与i位置匹配的串

           然后分情况讨论即可~