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.
# Anuneet Anand

T = int(input())

while T:
	s = input()
	D = {}
	for i in range(len(s)):
		if int(s[i]) in D:
			D[int(s[i])].append(i)
		else:
			D[int(s[i])]=[i]

	m = 0
	for i in D:
		m = max(m,len(D[i]))

	for i in D:
		for j in D:
			if i!=j:
				a = 1
				b = 0
				Z = []
				x = D[i][0]
				Z.append(x)
				while a<len(D[i])+1 and b<len(D[j]) and x<len(s):
					while b<len(D[j]) and D[j][b]<x:
						b = b + 1
					if b<len(D[j]):
						x = D[j][b]
						Z.append(x)
					else:
						x = len(s)+1
					while a<len(D[i]) and D[i][a]<x:
						a = a + 1
					if a<len(D[i]):
						x = D[i][a]
						Z.append(x)
					else:
						x = len(s)+1
				#print(i,j,Z)
				m = max(m,len(Z)-(len(Z)%2))
	print(len(s)-m)
	T = T - 1