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.
def fun(x):
    return x[0]

import sys
from collections import defaultdict
# t=1
t=int(input())
for i in range(t):
    n=list(input())
    lenn=len(n)
    # a=list(map(int,sys.stdin.readline().strip().split()))
    # a,b,c,d=list(sys.stdin.readline().strip().split())
    # n,k=list(map(int,sys.stdin.readline().strip().split()))
    
    
    d=defaultdict(int)
    # d['1']=1
    # print(d)
    
    for j in n:
        d[j]+=1
    # print(d)
    # print(max(d))
    
    freq=[]
    for j in range(10):
        freq.append([d[str(j)],j])
    # print(freq)
    
    freq.sort(key=fun,reverse=True)
    # print(freq)
    # x=set(n)
    # print(x.count())
    
    maxl=2
    maxl=max(maxl,freq[0][0])
    
    
    for j in range(10):
        for k in range(j+1,10):
            if(freq[j][0]+freq[k][0]<=maxl):
                break
            ll=0
            xx=str(freq[j][1])
            yy=str(freq[k][1])
            
            ii=0
            c=0
            while(ii<lenn):
                f1=0
                f2=0
                while(ii<lenn and not(n[ii]==yy)):
                    ii+=1
                if(ii==lenn):
                    break
                if(n[ii]==yy):
                    f1=1
                while(ii<lenn and not(n[ii]==xx)):
                    ii+=1
                if(ii==lenn):
                    break
                if(n[ii]==xx):
                    f2=1
                if(f1 and f2):
                    c+=2
            
            ll=c
            c=0
            ii=0
            c=0
            while(ii<lenn):
                f1=0
                f2=0
                while(ii<lenn and not(n[ii]==xx)):
                    ii+=1
                if(ii==lenn):
                    break
                if(n[ii]==xx):
                    f1=1
                while(ii<lenn and not(n[ii]==yy)):
                    ii+=1
                if(ii==lenn):
                    break
                if(n[ii]==yy):
                    f2=1
                if(f1 and f2):
                    c+=2
            ll=max(ll,c)
            # print(ll)
            maxl=max(maxl,ll)
    print(lenn-maxl)