// G. Hagopian doing Kattis Peragram

/*

Peragrams

Photo by Ross Beresford

Per recently learned about palindromes. Now he wants to tell us about it 
and also has more awesome scientific news to share with us.

"A palindrome is a word that is the same no matter whether you read it 
backward or forward", Per recently said in an interview. He continued: 
"For example, add is not a palindrome, because reading it backwards gives 
dda and it's actually not the same thing, you see. However, if we reorder 
the letters of the word, we can actually get a palindrome. Hence, we say 
that add is a Peragram, because it is an anagram of a palindrome".

Per gives us a more formal definition of Peragrams: "Like I said, if a word 
is an anagram of at least one palindrome, we call it a Peragram. And recall 
that an anagram of a word w contains exactly the same letters as w, possibly 
in a different order."

Task

Given a string, find the minimum number of letters you have to remove from 
it, so that the string becomes a Peragram. 

Input

Input consists of a string on a single line. The string will contain at 
least 1 and at most 1000 characters. The string will only contain lowercase 
letters a-z.

Output

Output should consist of a single integer on a single line, the minimum 
number of characters that have to be removed from the string to make it 
a Peragram.

Sample Input 1 	Sample Output 1
abc				2

Sample Input 2 	Sample Output 2
aab				0

Plan:
1.	Get the string
2.	sort the string
3.  count the number of times a letter recurs an odd number of times
	That's almost the answer!  The twist is that one letter may or may
	not appear an odd number of times.

Example:
"amanaplanacanalpanama"
when sorted becomes
 aaaaaaaaaacllmmnnnnpp
*/

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
	string s;
	int count{}, odds{};
	cin >> s;
	sort(s.begin(), s.end());
	int index{};
	char ch;
	while (index < s.size()) {
		ch = s[index];
		count = 1;
		while (s[++index] == ch) ++count;
		if (count % 2 == 1) ++odds;
	}
	if (odds > 0) cout << odds - 1;
	else cout << 0;
}