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 sys import stdin, stdout
from collections import Counter, defaultdict
pr=stdout.write
import heapq
raw_input  = stdin.readline
def ni():
    return int(raw_input())


def li():
    return list(map(int,raw_input().split()))


def pn(n):
    stdout.write(str(n)+'\n')


def pa(arr):
    pr(' '.join(map(str,arr))+'\n')

# fast read function for total integer input

def inp():
    # this function returns whole input of
    # space/line seperated integers 
    # Use Ctrl+D to flush stdin.
    return (map(int,stdin.read().split()))
    
range = xrange # not for python 3.0+

for t in range(ni()):
    s=raw_input().strip()
    d=defaultdict(list)
    n=len(s)
    for i in range(n):
        d[int(s[i])].append(i)
    ans=0
    ans1=0
    for i in range(10):
        ans=max(ans,len(d[i]))
        for j in range(10):
            arr=[]
            for i1 in d[i]:
                arr.append((i1,i))
            for i1 in d[j]:
                arr.append((i1,j))
            arr.sort()
            c=1
            for k in range(1,len(arr)):
                if arr[k][1]!=arr[k-1][1]:
                    c+=1
            ans=max(ans,c-c%2)
    pn(n-ans)
