description: A string s of length n, consisting of lowercase letters of the English alphabet, is given.
  
  You must choose some number k between 0 and n. Then, you select k characters of s and permute them however you want. In this process, the positions of the other n-k characters remain unchanged. You have to perform this operation exactly once.
  
  For example, if s="andrea", you can choose the k=4 characters "a_d_ea" and permute them into "d_e_aa" so that after the operation the string becomes "dneraa".
  
  Determine the minimum k so that it is possible to sort s alphabetically (that is, after the operation its characters appear in alphabetical order).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains one integer n (1 ≤ n ≤ 40) — the length of the string.
  
  The second line of each test case contains the string s. It is guaranteed that s contains only lowercase letters of the English alphabet.
  
  Output
  
  For each test case, output the minimum k that allows you to obtain a string sorted alphabetically, through the operation described above.
  
  Example
  
  Input
  
  
  4
  3
  lol
  10
  codeforces
  5
  aaaaa
  4
  dcba
  
  
  Output
  
  
  2
  6
  0
  4
  
  Note
  
  In the first test case, we can choose the k=2 characters "_ol" and rearrange them as "_lo" (so the resulting string is "llo"). It is not possible to sort the string choosing strictly less than 2 characters.
  
  In the second test case, one possible way to sort s is to consider the k=6 characters "_o__force_" and rearrange them as "_c__efoor_" (so the resulting string is "ccdeefoors"). One can show that it is not possible to sort the string choosing strictly less than 6 characters.
  
  In the third test case, string s is already sorted (so we can choose k=0 characters).
  
  In the fourth test case, we can choose all k=4 characters "dcba" and reverse the whole string (so the resulting string is "abcd").
solution: #include <bits/stdc++.h>
  using namespace std;
  using namespace chrono;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long expo(long long a, long long b, long long mod) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  void extendgcd(long long a, long long b, long long *v) {
    if (b == 0) {
      v[0] = 1;
      v[1] = 0;
      v[2] = a;
      return;
    }
    extendgcd(b, a % b, v);
    long long x = v[1];
    v[1] = v[0] - v[1] * (a / b);
    v[0] = x;
    return;
  }
  long long mminv(long long a, long long b) {
    long long arr[3];
    extendgcd(a, b, arr);
    return arr[0];
  }
  long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
  bool revsort(long long a, long long b) { return a > b; }
  void swap(int &x, int &y) {
    int temp = x;
    x = y;
    y = temp;
  }
  long long combination(long long n, long long r, long long m, long long *fact,
                        long long *ifact) {
    long long val1 = fact[n];
    long long val2 = ifact[n - r];
    long long val3 = ifact[r];
    return (((val1 * val2) % m) * val3) % m;
  }
  void google(int t) { cout << "Case #" << t << ": "; }
  vector<long long> sieve(int n) {
    int *arr = new int[n + 1]();
    vector<long long> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  long long mod_add(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mod_mul(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long mod_div(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
  }
  long long phin(long long n) {
    long long number = n;
    if (n % 2 == 0) {
      number /= 2;
      while (n % 2 == 0) n /= 2;
    }
    for (long long i = 3; i <= sqrt(n); i += 2) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        number = (number / i * (i - 1));
      }
    }
    if (n > 1) number = (number / n * (n - 1));
    return number;
  }
  void solve() {
    long long n;
    cin >> n;
    string s;
    cin >> s;
    string store = s;
    sort(s.begin(), s.end());
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      if (store[i] != s[i]) {
        ans++;
      }
    }
    cout << ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      solve();
      cout << "\n";
    }
  }
  