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.
import sys
t = sys.stdin.readline()
while t:
    t = int(t)
    s = input()
    n = len(s)
    ans = n
    for i in range(100):
        if i < 10:
            a,b = "0",str(i)
        else:
            b = str(i%10)
            a = str(i//10)
        
        j = 0
        flag = True
        gd_str = 0
        while j < n:
            if flag:
                while j < n and s[j] != a:
                    j += 1
                flag = False
            else:
                while j < n and s[j] != b:
                    j += 1
                    continue
                if j < n and s[j] == b:
                    gd_str += 2
                
                flag = True
            
            j += 1
        ans = min(ans, n-gd_str)
    hm = {}
    for i in range(n):
        try:
            hm[s[i]] += 1
        except:
            hm[s[i]] = 1
    print(min(n-max(hm.values()) , ans))
    t -= 1
