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.
import sys
input = sys.stdin.readline

def name(arr):
  # h = {}
  patterns = {}
  highest = 0
  count = 0 
  for i in range(10):
    for j in range(10):
      # if i != j:
      first = 0
      found = False
      for char in arr[:-1]:
        if i == j:
          # if count % 2 == 0:
          if char == str(i):
            count += 1
        else:
          if first != 1:
            if char == str(i):
              first = 1
          elif first == 1:
            if char == str(j):
              first = 0
              found = True
          if found:
            count += 2
            found = False
      if count > highest:
        highest = count
        
      count = 0
  return len(arr) -1- highest
  

def solution():
  arr = input()
  print(name(arr))

t = int(input());
for i in range(0, t):
  solution()
  
