# Anagram

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#
#   Sid is obsessed with reading short stories. Being a CS student, he is doing some interesting frequency analysis
#   with the books. He chooses strings S1 and S2 in such a way that |len(S1) - len(S2)| <= 1.
#   Your task is to help him find the minimum number of characters of the first string he needs to change to enable
#   him to make it an anagram of the second string.
#   Note: A word x is an anagram of another word y if we can produce y by rearranging the letters of x.
#
#   Input Format
#   The first line will contain an integer, T, representing the number of test cases. Each test case will contain a
#   string having length len(S1) + len(S2), which will be concatenation of both the strings described above in the
#   problem.The given string will contain only characters from a to z.
#
#   Output Format
#   An integer corresponding to each test case is printed in a different line, i.e. the number of changes required
#   for each test case. Print -1 if it is not possible.
#
#   Constraints
#   1 <= T <= 100
#   1 <= len(S1) + len(S2) <= 10^4
#
#   Sample Input
#   6
#   aaabbb
#   ab
#   abc
#   mnop
#   xyyx
#   xaxbbbxx
#
#   Sample Output
#   3
#   1
#   -1
#   2
#   0
#   1
#
#   Explanation
#   Test Case #01: We have to replace all three characters from the first string to make both of strings anagram.
#   Here, S1 = "aaa" and S2= "bbb". So the solution is to replace all character 'a' in string a with character 'b'.
#   Test Case #02: You have to replace 'a' with 'b', which will generate "bb".
#   Test Case #03: It is not possible for two strings of unequal length to be anagram for each other.
#   Test Case #04: We have to replace both the characters of first string ("mn") to make it anagram of other one.
#   Test Case #05: S1 and S2 are already anagram to each other.
#   Test Case #06: Here S1 = "xaxb" and S2 = "bbxx". He had to replace 'a' from S1 with 'b' so that S1 = "xbxb"
#   and we can rearrange its letter to "bbxx" in order to get S2.
#
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n = int(input().strip())

for wallCare in range(n):
    mantra = input().strip()

    if len(mantra)%2 != 0 :
        print(-1)
    else:
        #break
        nashesi = len(mantra)//2
        banA = list(mantra[0: nashesi])
        supriyaB = list(mantra[nashesi :len(mantra)])
        #sort
        banA.sort()
        supriyaB.sort()

        #check
        misYada = 0
        for bhaiRok in range(len(banA)):
            if banA[bhaiRok] == supriyaB[bhaiRok]:
                pass
            else:
                misYada += 1
        print(misYada)
        misYada = 0
