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 math

T = int(input())

for i in range(T):
  #n = int(input())
  #n,k,z = map(int, input().split())
  #a = list(map(int,input().split()))
  a = input()
  d = False
  f = {}
  q = list('0123456789')
  n = len(a)


  first = [1]*100
  count = [0]*100

  for i in a:
    j = int(i)
    #print(j,'-')
    #print(first[25])
    for k in range(10):
      if first[j*10+k]:
        count[j*10+k] += 1
        first[j*10+k] ^= 1
      if not first[k*10+j]:
        count[10*k+j] += 1
        first[10*k+j] ^= 1


  ans = 0
  for i in range(100):
    if i%10 == i//10:
      ans = max(ans,count[i]//2)
    else:
      ans = max(ans,count[i]-count[i]%2)
  
  print(n-ans)
