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;
  template <typename T>
  using minpq = priority_queue<T, vector<T>, greater_equal<T>>;
  long long MOD = 1e9 + 7;
  template <typename T>
  istream &operator>>(istream &is, vector<T> &arr) {
    typename vector<T>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      is >> *a;
    }
    return is;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> &arr) {
    typename vector<T>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      os << *a << " ";
    }
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, pair<T, U> &p) {
    os << p.first << " " << p.second << '\n';
    return os;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &os, vector<pair<T1, T2>> &arr) {
    typename vector<pair<T1, T2>>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      os << *a;
    }
    return os;
  }
  long long mul(long long a, long long b, long long md = MOD) {
    return (a % md * b % md) % md;
  }
  long long add(long long a, long long b, long long md = MOD) {
    long long ans = a + b;
    if (ans >= md) ans -= md;
    return ans;
  }
  bool valid(int i, int j, int n, int m) {
    return i >= 0 && i < n && j >= 0 && j < m;
  }
  vector<pair<int, int>> dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
  long long binPow(long long x, long long y) {
    long long ans = 1;
    while (y > 0) {
      if (y % 2 == 1) ans = mul(ans, x);
      x = mul(x, x);
      y /= 2;
    }
    return ans;
  }
  long long modInverse(long long x) { return binPow(x, MOD - 2); }
  void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    string t(s.begin(), s.end());
    sort(t.begin(), t.end());
    for (int i = 0; i < int(((int)s.size())); i++) {
      if (s[i] == t[i]) n--;
    }
    cout << n << endl;
  }
  int main() {
    (ios_base::sync_with_stdio(false), cin.tie(0));
    int t;
    cin >> t;
    while (t--) solve();
  }
  