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.
t=int(input())
for i in range(t):
  s=input()
  v=len(s)
  ansf=0
  bq=[0]*10
  l=['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
  for j in range(len(s)):
    bq[int(s[j])]+=1
  ansf=max(bq)
  for j in range(10):
    for k in range(10):
      if(j!=k):
        ans=0
        w=0
        while(w<v):
          p=0
          q=0
          while(w<v and s[w]!=l[j]):
            w+=1
            p=1
          if(p==1):
            w+=1
            #ans+=1
            while(w<v and s[w]!=l[k]):
              w+=1
              q=1
            if(q==1 and w<v):
              ans+=2
              #print("P")
              w+=1
            else:
              if(w<v):
                #print("P")
                ans+=2
              w+=1
          else:
            if(w<v):
              w+=1
              while(w<v and s[w]!=l[k]):
                w+=1
                q=1
              if(q==1 and w<v):
                ans+=2
                w+=1
              else:
                if(w<v):
                  #print("P")
                  ans+=2
                w+=1
            #w+=1
        ansf=max(ans,ansf)
        #print(ansf,j,k)
  print(v-ansf)