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 
from decimal import *
def I(): return int(sys.stdin.readline())
def IL(): return list(map(int,sys.stdin.readline().strip().split()))
def ILS(): return list(map(str,sys.stdin.readline().strip().split()))

def solve():
	tt = I()
	for ii in range(tt):
		s = input()
		ss = [int(x) for x in s]
		n = len(ss)
		ans = n-2
		count = [0]*10
		for i in range(n):
			count[ss[i]]+=1
		for i in range(0,10):
			ans = min(ans,n-count[i])
		for f in range(0,10):
			for k in range(0,10):
				res = 0
				for i in range(n):
					if res%2 == 0 and f == ss[i] : res += 1
					elif res %2 ==1 and k == ss[i] :res+=1
				if res%2 == 1 : continue
				ans = min(ans,n-res)
		
		print(ans)
	


		pass
		

solve()