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.
for _ in range(0, int(input())):
	l = list(input())

	d={}
	for i in range(0, len(l)):
		if int(l[i]) in d:
			d[int(l[i])].append(i)
		else:
			d.update({int(l[i]): [i]})
	# print(d)  

	odd_res = 0
	even_res=999999999
	for i in d:
		odd_res = max(len(d[i]), odd_res)
	odd_res = len(l) - odd_res
	if odd_res==len(l)-1:
		odd_res-=1
	# print(odd_res, "odd_res")

	for i in range(0, 10):
		for j in range(0, 10):
			if i!=j:
				temp=[]
				for m in range(0, len(l)):
					if int(l[m]) in [i, j]:
						if temp!=[] and temp[-1]==l[m]:
							pass
						else:
							temp.append(l[m])
				# print(temp)
				if temp!=[]:
					flag5=0
					tem=0
					for m in range(0, len(temp)-1):
						if temp[m]==temp[m+1]:
							tem+=1
					if flag5==0:
						if len(temp)%2==0:
							even_res = min(even_res, len(l)-(len(temp)))
						else:
							even_res = min(even_res, len(l)-(len(temp))+1)
	# print(even_res)
	print(min(odd_res, even_res))