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()
    ans = 0 
    dic = [0]*10
    mx = 0 
    for i in s:
        dic[int (i)] += 1
        if mx <dic[int (i)]:
            mx = dic[int (i)]
    ans = len(s) - mx 

    s1  = '0'
    s2 = '0'
    flag = 0
    while int(s1) <= 9 and int(s2) <= 9:
        for i in s:
            if (flag &1 and s2 == i) or (flag%2 == 0 and s1 == i):
                flag += 1
        
        if flag %2 == 0 :
            # print(flag//2)
            ans = min (len(s) - flag , ans)
        v = int (s1 + s2) +1
        s1 = str(v//10)
        s2 = str(v%10)
        flag = 0
    print(ans)
