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
import math
def II():
	return int(sys.stdin.readline())

def LI():
	return list(map(int, sys.stdin.readline().split()))

def MI():
	return map(int, sys.stdin.readline().split())

def SI():
	return sys.stdin.readline().strip()
t = II()
for q in range(t):
	s = list(SI())
	ans = 2
	d = [[] for i in range(10)]
	dp = [0]*10
	for i in range(len(s)):
		d[int(s[i])].append(i)
		dp[int(s[i])]+=1
	ans = max(ans,max(dp))
	for i in range(10):
		if len(d[i]) == 0:
			continue
		for j in range(i+1,10):
			count = 1
			a = d[i][:]
			b = d[j][:]
			if len(b) == 0:
				continue
			l = 0
			l2 = 0
			temp = -1
			if a[0]<b[0]:
				temp = a[0]
				boo = False
				end = True
			else:
				temp = b[0]
				boo = True
				end = False
			while l<dp[i] and l2<dp[j]:
				if boo:
					while l<dp[i] and a[l]<temp:
						l+=1
					if l<dp[i]:
						temp = a[l]
						count+=1
						boo = not boo
					else:
						break
				else:
					while l2<dp[j] and b[l2]<temp:
						l2+=1
					if l2<dp[j]:
						temp = b[l2]
						count+=1
						boo = not boo
					else:
						break
			if end != boo:
				count-=1
			ans = max(ans,count)
	print(len(s)-ans)




