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.
for _ in range(int(input())):
    s=input()
    n=len(s)
    ans=n-2
    g={i:[]for i in '0123456789'}
    for i in range(n):g[s[i]].append(i)
    for a in'0123456789':
        for b in'0123456789':
            cs=0
            i=j=0
            aa=g[a]
            bb=g[b]
            ans=min(ans,n-len(aa))
            while i<len(aa):
                while j<len(bb)and aa[i]>=bb[j]:j+=1
                if j<len(bb):
                    while i<len(aa)and aa[i]<=bb[j]:i+=1
                else:break
                cs+=1
            ans=min(ans,n-cs*2)
    print(ans)