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 __future__ import print_function, division
from sys import stdin, stdout
from fractions import gcd
from math import *
from collections import *
from operator import itemgetter, mul
from functools import reduce
from copy import copy

rstr = lambda: stdin.readline().strip()
rstrs = lambda: [str(x) for x in stdin.readline().split()]
rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]
arr_2d = lambda n: [rints() for i in range(n)]
out = []

for _ in range(int(input())):
    s, ans = rstr(), float('inf')

    for j in range(10):
        for k in range(10):
            dig = [str(j), str(k)]
            cur, tem, l = 0, 0, 0

            for i in s:
                if i == dig[cur]:
                    cur ^= 1
                    l += 1
                else:
                    tem += 1

            ans = min(ans, tem + (l % 2) if j != k else tem)

    out.append(ans)

print('\n'.join(map(str, out)))
