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 math import *
from collections import *
from random import *
from bisect import *
import sys
input=sys.stdin.readline
t=int(input())
while(t):
    t-=1
    s=input().rstrip('\n')
    f=Counter(s)
    x=len(s)
    r=float('inf')
    if(len(f)==1):
        print(0)
        continue
    d={i:[] for i in range(10)}
    for i in range(len(s)):
        try:
            d[int(s[i])].append(i)
        except:
            d[s[i]]=[i]
    for i in range(10):
        for j in range(10):
            if(i==j):
                r=min(r,len(s)-len(d[i]))
                continue
            c=1
            x,y=0,0
            try:
                fi=d[i][0]
            except:
                continue
            tu=0
            while(x<len(d[i]) or y<len(d[j])):
                if(tu%2 ^1):
                    if(y>=len(d[j])):
                        break
                    if(d[j][y]>fi):
                        c+=1
                        y+=1
                        se=d[j][y-1]
                        tu+=1
                    else:
                        y+=1
                        continue
                else:
                    if(x>=len(d[i])):
                        break
                    if(d[i][x]>se):
                        c+=1
                        tu+=1
                        x+=1
                        fi=d[i][x-1]
                    else:
                        x+=1
                        continue
            r=min(r,len(s)-(c-(c%2)))
    print(r)
                    
            
            
        
    
