Let's call left cyclic shift of some string t_1 t_2 t_3 ... t_{n - 1} t_n as string t_2 t_3 ... t_{n - 1} t_n t_1.

Analogically, let's call right cyclic shift of string t as string t_n t_1 t_2 t_3 ... t_{n - 1}.

Let's say string t is good if its left cyclic shift is equal to its right cyclic shift.

You are given string s which consists of digits 0–9.

What is the minimum number of characters you need to erase from s to make it good?

Input

The first line contains single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Next t lines contains test cases — one per line. The first and only line of each test case contains string s (2 ≤ |s| ≤ 2 ⋅ 10^5). Each character s_i is digit 0–9.

It's guaranteed that the total length of strings doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the minimum number of characters you need to erase from s to make it good.

Example

Input


3
95831
100120013
252525252525


Output


3
5
0

Note

In the first test case, you can erase any 3 characters, for example, the 1-st, the 3-rd, and the 4-th. You'll get string 51 and it is good.

In the second test case, we can erase all characters except 0: the remaining string is 0000 and it's good.

In the third test case, the given string s is already good.
for _ in range(int(input())):
  s=input()
  we=len(s)
  d={}
  l=[]
  for i in s:
    if i not in d:
      d[i]=1
    else:
      d[i]+=1
  
  for i in d:
    l.append(d[i])
  
  l_pair=[]
  for i in d:
    for j in d:
      if i!=j:
        ss=i+j
        l_pair.append(ss)

  for i in range(len(l_pair)):
    cnt=0
    flag=0
    for j in range(we):
      if flag==0 and s[j]==l_pair[i][0]:
        flag=1
      elif flag==1 and s[j]==l_pair[i][1]:
        cnt+=1
        flag=0
    l.append(cnt*2)
  
  an=max(max(l),2)
  print(we-an)
  
 # print(l_pair)
  #print(l,d)