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: # author : Tapan Goyal
  #           MNIT Jaipur
  
  import math
  import bisect
  import itertools
  import sys
  I=lambda : sys.stdin.readline()
  one=lambda : int(I())
  more=lambda : map(int,I().split())
  linput=lambda : list(more())
  
  mod=10**9 +7
  inf=10**18 + 1 
  
  '''fact=[1]*100001
  ifact=[1]*100001
  for i in range(1,100001):
      fact[i]=((fact[i-1])*i)%mod
      ifact[i]=((ifact[i-1])*pow(i,mod-2,mod))%mod
  def ncr(n,r):
      return (((fact[n]*ifact[n-r])%mod)*ifact[r])%mod
  def npr(n,r):
      return (((fact[n]*ifact[n-r])%mod))
      '''
  
  
  def merge(a,b):
      i=0;j=0
      c=0
      ans=[]
      while i<len(a) and j<len(b):
          if a[i]<b[j]:
              ans.append(a[i])
              i+=1
          else:
              ans.append(b[j])
              c+=len(a)-i
              j+=1
      ans+=a[i:]
      ans+=b[j:]
      return ans,c
  def mergesort(a):
      if len(a)==1:
          return a,0
      mid=len(a)//2   
      left,left_inversion=mergesort(a[:mid])
      right,right_inversion=mergesort(a[mid:])
      m,c=merge(left,right)
      c+=(left_inversion+right_inversion)
      return m,c
      
  def is_prime(num):
      if num == 1: return False
      if num == 2: return True
      if num == 3: return True
      if num%2 == 0: return False
      if num%3 == 0: return False
      t = 5
      a = 2
      while t <= int(math.sqrt(num)):
          if num%t == 0: return False
          t += a
          a = 6 - a
      return True
  
    
  def ceil(a,b):
      return (a+b-1)//b
      
  
  
  #/////////////////////////////////////////////////////////////////////////////////////////////////
  if __name__ == "__main__":
      for _ in range(one()):
          def solve():
              n=one()
              ans=0
              s=list(input())
              p=sorted(s)
              for i in range(n):
                  if s[i]!=p[i]:
                      ans+=1
              print(ans)        
          solve()    