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.
t = int(input())
for i in range(t):
	a = input()
	b = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
	k = 0
	for j in range(10):
		for l in range(10):
			q = str(b[j])
			w = str(b[l])
			e = 0
			r = 0
			d = 0
			for f in a:
				d += 1
				if f == q and e == r:
					e += 1
				elif f == w and e != r:
					r += 1
			if q == w:
				if e + r > k:
					k = e + r
			else:
				if e + r - (e + r) % 2 > k:
					k = e + r - (e + r) % 2
	print(d - k)