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.
def alt(a,i,j):
	c=0
	now=i
	for k in range(len(a)):
		if(a[k]==i and now==i):
			now=j
		elif(a[k]==j and now==j):
			now=i
			c+=1
	return c*2
def main():
	for _ in range(int(input())):
		a=list(input())
		x=[i for i in range(10)]
		m=0
		for i in range(10):
			for j in range(10):
				if(i!=j):
					m=max(m,alt(a,str(i),str(j)))
				else:
					m=max(m,a.count(str(i)))
		print(len(a)-m)
main()