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 uint = unsigned int;
  using ll = long long;
  using ull = unsigned long long;
  using db = double;
  using ld = long double;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using pdd = pair<double, double>;
  using vint = vector<int>;
  using vpii = vector<pii>;
  pii operator+(pii a, pii b) { return {a.first + b.first, a.second + b.second}; }
  pll operator+(pll a, pll b) { return {a.first + b.first, a.second + b.second}; }
  namespace IO {
  char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 24], *O = obuf;
  inline ll read() {
    ll x = 0;
    bool sign = 0;
    char s =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
             ? EOF
             : *p1++);
    while (!isdigit(s))
      sign |= s == '-',
          s = (p1 == p2 &&
                       (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
                   ? EOF
                   : *p1++);
    while (isdigit(s))
      x = (x << 1) + (x << 3) + (s - '0'),
      s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
               ? EOF
               : *p1++);
    return sign ? -x : x;
  }
  inline void print(ll x) {
    if (x < 0)
      (*O++ = '-'), print(-x);
    else {
      if (x > 9) print(x / 10);
      (*O++ = x % 10 + '0');
    }
  }
  }  // namespace IO
  const int mod = 998244353;
  const long double Pi = acos(-1);
  namespace math {
  const int MAXN = 2e6 + 5;
  ll ksm(ll a, ll b, ll p) {
    ll s = 1, m = a;
    while (b) {
      if (b & 1) s = s * m % p;
      m = m * m % p, b >>= 1;
    }
    return s;
  }
  ll ksm(ll a, ll b) {
    ll s = 1, m = a;
    while (b) {
      if (b & 1) s = s * m % mod;
      m = m * m % mod, b >>= 1;
    }
    return s;
  }
  ll inv(ll x) { return ksm(x, mod - 2); }
  ll fc[MAXN], ifc[MAXN];
  void init(int n) {
    assert(n < MAXN);
    fc[0] = 1;
    for (int i = 1; i <= n; i++) fc[i] = fc[i - 1] * i % mod;
    ifc[n] = inv(fc[n]);
    for (int i = n - 1; i >= 0; i--) ifc[i] = ifc[i + 1] * (i + 1) % mod;
  }
  ll bin(ll n, ll m) { return fc[n] * ifc[m] % mod * ifc[n - m] % mod; }
  }  // namespace math
  mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
  using namespace IO;
  const int N = 1e3 + 5;
  void solve() {
    int n;
    string s, t;
    cin >> n >> s, t = s;
    sort(t.begin(), t.end());
    int cnt = 0;
    for (int i = 0; i < n; i++) cnt += s[i] != t[i];
    cout << cnt << endl;
  }
  int main() {
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  