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.
from collections import Counter

def f(s):
 m=1000000; u=Counter(s)
 for c in '0123456789': m=min(m,len(s)-u[c])
 t={c:[] for c in '0123456789'}
 for i in xrange(len(s)): t[s[i]].append(i)
 for i in range(10):
  i=str(i)
  if not(t[i]): continue
  for j in xrange(int(i)+1,10):
   j=str(j)
   if not(t[j]): continue
   a=0; b=0; r=(t[i][a]>t[j][b]); v=1
   while a<len(t[i]) and b<len(t[j]):
    if r==0:
     v+=1
     while a<len(t[i]) and t[i][a]<t[j][b]: a+=1
     r=1
    else:
     v+=1
     while b<len(t[j]) and t[j][b]<t[i][a]: b+=1
     r=0
   m=min(m,len(s)-(v/2)*2)
 return m

for t in xrange(int(raw_input())):
 s=raw_input().strip()
 print f(s)
