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()
    f = {}
 
    for i in range(len(s)):
        if int(s[i]) not in f:
            f[int(s[i])] = 1
        else:
            f[int(s[i])] += 1
 
    m = max(f.values())
    t = len(f)
    
            
    c,a = 0,0
    #print(f)
    
    if m == 1:
        print(len(s)-2)
    elif m == len(s):
        print(0)
    else:
        l = len(s)
        for i in range(10):
            for j in range(10):
                f = 1
                for p in range(l):
                    if int(s[p]) == i and f == 1:
                        c += 1
                        f = 0
                    elif int(s[p]) == j and f == 0:
                        c += 1
                        f = 1
 
                if i != j and c%2 == 1 :
                    c -= 1
                a = max(a,c)
                c = 0
 
 
        print(l-a)