// Arup Guha
// 6/1/2013
// Solution to 2013 KTH Challenge Problem B: Peragrams

import java.util.*;

public class peragrams {

	public static void main(String[] args) {

		Scanner stdin = new Scanner(System.in);
		boolean[] odd = new boolean[26];
		String s = stdin.next();

		// Keep track of parity of frequencies of each letter.
		for (int i=0; i<s.length(); i++)
			odd[s.charAt(i)-'a'] = !odd[s.charAt(i)-'a'];

		// cnt is how many letters have a odd frequency.
		int cnt = 0;
		for (int i=0; i<26; i++)
			if (odd[i])
				cnt++;

		// We can have at most one letter of odd frequency.
		System.out.println(Math.max(0, cnt-1));

	}
}