int anagram(string s) {
    if(s.size() % 2 == 1) return -1;
    map<char, int> mp;
    int ans = 0;
    for(int i = 0; i < s.size() / 2; i++) mp[s[i]]++;
    for(int i = s.size() / 2; i < s.size(); i++){
        if(mp[s[i]] != 0) mp[s[i]]--;
        else ans++;
    }
    return ans;
}

sub anagram {
    my $s = shift;
    my %h1;
    my $cnt = 0;

    if (length($s) % 2) {
        return -1;
    }

    my $s1 = substr($s, 0, length($s) / 2);
    my @s2 = split("", substr($s, length($s) / 2, length($s)));

    
    foreach (split("", $s1)) {
        $h1{$_} += 1;
    }

    foreach my $k (@s2) {
        if (defined($h1{$k}) && $h1{$k} != 0) {
            $h1{$k} -= 1;            
        }
        else {
            $cnt++;
        }        
    }

    return $cnt;
}

function anagram(s: string): number {
    // 0. check s length is odd, so it cann't be split, return -1
    if (s.length % 2 != 0) return -1

    // 1. split [s] into left part [l] and right part [r], define [m]&[o] to count
    let l = s.substring(0, s.length / 2)
    let r = s.substring(s.length / 2)
    let m = new Map<string, number>()
    let o = 0

    // 2. count [l] into [m], check [r], count++
    for (let c of l) m.set(c, (m.get(c) || 0) + 1)
    for (let c of r) {
        if (m.has(c) && m.get(c) > 0) m.set(c, m.get(c) - 1)
        else o++
    }

    // 3. return [o]
    return o
}

from collections import Counter

def anagram(s):
    n = len(s)
    if n % 2 == 1:
        return -1  
    a = s[:n//2]
    b = s[n//2:]
    freq_a = Counter(a)
    freq_b = Counter(b)
    changes = 0
    for char in freq_a:
        if freq_a[char] > freq_b.get(char, 0):
            changes += freq_a[char] - freq_b.get(char, 0)
    return changes