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 debug import debug
import sys
input = sys.stdin.readline
inf = int(1e10)
# from math import ceil, log2

t = int(input())
while t:
	t-=1
	s =	input().strip()
	n = len(s)
	d = dict()
	for i in s:
		if i in d: d[i] += 1
		else: d[i] = 1
	m1 = max(d.values())
	m2 = 0
	for i in range(10):
		for j in range(10):
			if i!=j:
				f = 0
				l = 0
				for k in s:
					if f == 0 and k == str(i): f = 1; l+=1
					elif f == 1 and k == str(j): f = 0; l+=1
				m2 = max(l, m2)
	print(n-max(m1, (m2>>1)<<1))
