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 ans(s,x):
    ana=0
    bo=0
    for i in range(len(x)):
        if(bo==0):
            if x[i]==s[0]:
               ana+=1
               bo=1
               continue
        if(bo==1):
            if x[i]==s[1]:
                ana+=1
                bo=0
    
    
    return ana
def solve(x):
    a=["00","01","02","03","04","05","06","07","08","09","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31","32","33","34","35","36","37","38","39","40","41","42","43","44","45","46","47","48","49","50","51","52","53","54","55","56","57","58","59","60","61","62","63","64","65","66","67","68","69","70","71","72","73","74","75","76","77","78","79","80","81","82","83","84","85","86","87","88","89","90","91","92","93","94","95","96","97","98","99"]
    res=0
    for i in range(len(a)):
        aaa=ans(a[i],x)
        if(aaa%2 and a[i][0]!=a[i][1]):
            aaa-=1
        res=max(res,aaa)
    print(len(x)-res)
    
# cook your dish here
for i in range(int(input())):
    x=(input())
    y=set(x)
    if(len(x)==len(y)):
        print(len(x)-2)
        continue
    else:
        solve(x)
        