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 _ in range(t):
	s = input()
	n = len(s)
	ans = [0]*10
	for i in range(n):
		ans[ord(s[i]) - ord('0')] = ans[ord(s[i]) - ord('0')] + 1
	ans.sort()
	mx = ans[9]
	for i in range(10):
		for j in range(10):
			c = 0
			if(j==i):
				continue
			b = 0
			for k in range(n):
				if(b==0 and (ord(s[k]) - ord('0')) == i):
					b = 1
					c = c + 1
				elif(b == 1 and (ord(s[k]) - ord('0')) == j):
					b = 0
					c = c+1
				mx = max(c - c%2, mx)
	print(n - max(mx, ans[9]))