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.
# Author: S Mahesh Raju
# Username: maheshraju2020
# Date: 29/07/2020

from sys import stdin,stdout
from math import gcd, ceil, sqrt
from collections import Counter
from bisect import bisect_left, bisect_right
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007

def make(a, b):
    i = j = 0
    arr = []
    while i < len(a) and j < len(b):
        if a[i] < b[j]:
            if not len(arr) or arr[-1] != 'a':
                arr.append('a')
            i += 1
        else:
            if not len(arr) or arr[-1] != 'b':
                arr.append('b')
            j += 1
    
    if i < len(a) and arr[-1] != 'a':
        arr.append('a')
    if j < len(b) and arr[-1] != 'b':
        arr.append('b')
    if len(arr) % 2 != 0:
        arr.pop()
    return len(arr)
        

tc = ii1()
for _ in range(tc):
    s = is1()
    d = Counter(s)
    res = 0
    for i in d:
        res = max(res, d[i])
    res = min(len(s) - 2, len(s) - res)

    d = {}
    for i in range(len(s)):
        d.setdefault(int(s[i]), []).append(i)

    for i in d:
        for j in d:
            if i != j:
                t = len(s) - make(d[i], d[j])
                res = min(res, t)
    print(res)