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 collections import defaultdict,Counter
import string
import random
import time



dig = string.digits

def tree():
    return defaultdict(tree)

t = int(input())
# s = "".join([random.choice(dig) for i in range(2*10**5)])
# time_start=time.time()
for _ in range(t):
    s = raw_input()
    
    num_dict = Counter(s)
    res = max(num_dict.values())
    f = tree()
    g = tree()
    # f,g = {},{}

    dig_dict = defaultdict(list)
    for i in dig:
        for j in dig:
            if i!=j and num_dict[i]+num_dict[j]>res:
                # if i not in f:
                #     f[i] = {}
                #     g[i] = {}
                #     dig_dict[i] = []
                f[i][j] = -1
                g[i][j] = 0
                dig_dict[i].append(j)

    
    for i in s:
        if i not in dig_dict:
            continue
        for j in dig_dict[i]:
            f[i][j] = max(f[i][j], g[i][j]+1)
            if f[j][i]>0:
                g[j][i] = max(g[j][i], f[j][i]+1)

    if dig_dict:
        tmp = max([g[i][j] for i,vlist in dig_dict.items() for j in vlist])
    else:
        tmp = 0
    print(len(s) - max(tmp, res))
        

# time_end=time.time()
# print('time cost',time_end-time_start,'s')