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 io
import os
import sys
from collections import deque, defaultdict


def _input():
    x = sys.stdin.readline()
    x = x.replace('\r', '')
    x = x.replace('\n', '')
    return x


if 'STDINPUT' not in os.environ:
    input = _input
else:
    print('Standard input')


def solve():
    x = [int(i) for i in input()]
    pattern = []
    for i in range(10):
        for j in range(10):
            pattern.append([i, j])
    ans = 10**10
    cans = 0
    g = 0
    for i in pattern:
        for j in x:
            if j == i[g % 2]:
                cans += 1
                g += 1
        ans = min(ans, len(x) - (cans - cans % 2 if not i[0] == i[1] else cans))
        cans = 0
        g = 0
    print(ans)





test_cases = True
if test_cases:
    _q = int(input())
    for _ in range(_q):
        solve()
else:
    solve()
