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 sys
input = sys.stdin.readline
def inp():
	return(int(input()))
def inlt():
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(s[:len(s)-1])
def invr():
	return(map(int,input().split()))
T = inp()
for t in range(T):
	S = insr()
	if len(S) < 3:
		print(0)
	else:
		maxx = 0
		for i in range(100):
			curr = str(i)
			if i < 10:
				curr = "0"+curr
			fl = True
			tmp = 0
			for j in range(len(S)):
				if fl:
					if S[j] == curr[0]:
						fl = False
				else:
					if S[j] == curr[1]:
						fl = True
						tmp +=2
			maxx = max(tmp,maxx)
		for i in range(10):
			tmp = S.count(str(i))
			maxx = max(tmp,maxx)
		print(len(S)-maxx)