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 # for PyPy2
from collections import Counter, OrderedDict
from itertools import permutations as perm
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import math
 
g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")

t, = gil()

def cnt(chars, s):
	p = 0
	cnt = 0
	for ch in s:
		if ch == chars[p]:
			p ^= 1
			cnt += 1
	# print(s, chars, (cnt - (cnt&1)))
	return (cnt - (cnt&1))

# print(cnt(["2", "5"], "252525252525"))
for _ in range(t):
	# print()
	s = g()
	n = len(s)
	b = Counter(s)
	ans = 0
	for ch in b:
		ans = max(ans, b[ch])

	# print(n -ans, end=" ")

	for i in range(10):
		for j in range(i+1, 10):
			ans = max(ans, cnt([str(i), str(j)], s), cnt([str(j), str(i)], s))

	print(n-ans)

