human
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").
# aut ... Write me a Python code to determine the minimum k so that it is possible to sort s alphabetically (that is, after the operation its characters appear in alphabetical order)?
assistant
#Here is Python code to determine the minimum k so that it is possible to sort s alphabetically (that is, after the operation its characters appear in alphabetical order).
hor : 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()