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 now on solve easy one first
t=int(input())
def subseq(a,b,s):
    ans=0
    i=0
    flag=0
    while i<len(s):
        if flag==0:
            if s[i]==a:
                ans+=1
                flag+=1
        else:
            if s[i]==b:
                ans+=1
                flag+=1
        flag%=2
        i+=1
    if ans%2==0:
        return ans
    elif a==b :
        return ans
    else:
        return ans-1
            
def solve(a):
    ans=10**9
    for i in range(10):
        for j in range(10):
            ans=min(ans,len(a)-subseq(str(i),str(j),a))
           
    print(ans)
while t:
    t-=1
    a=input()
    solve(a)
