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
import math
from collections import defaultdict,Counter

# input=sys.stdin.readline
# def print(x):
#     sys.stdout.write(str(x)+"\n")

# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')

# m=pow(10,9)+7
t=int(input())
for i in range(t):
    s=input()
    n=len(s)
    ma=0
    mi=10
    for j in s:
        ma=max(ma,int(j))
        mi=min(mi,int(j))
    # v=[[0]*10 for j in range(10)]
    # v=set()
    ans=0
    for j in range(mi,ma+1):
        for k in range(mi,ma+1):
            if j==k:
                c=s.count(str(j))
            else:
                c=0
                flag=0
                for ch in s:
                    # v.add(ch)
                    if flag==0 and int(ch)==j:
                        flag=1
                    elif flag==1 and int(ch)==k:
                        flag=0
                        c+=2
            ans=max(ans,c)
    # if n&1 and n-ans==1 and len(v)==1:
    #     ans=n
    print(n-ans)