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,math as mt
import heapq as hp
import math as mt
import collections as cc
I=lambda:list(map(int,input().split()))
def ch(s,a,b):
	aa=1
	bb=0
	cnt=0
	for i in range(len(s)):
		if s[i]==a and aa:
			aa=0
			bb=1
		elif s[i]==b and bb:
			bb=0
			aa=1
			cnt+=1
	return len(s)-(cnt*2)
for tc in range(int(input())):
	s=list(input())
	ans=mt.inf
	n=len(s)
	f=cc.Counter(s)
	for i in f:
		ans=min(ans,n-f[i])
	for i in range(10):
		for j in range(10):
			i=str(i)
			j=str(j)
			ans=min(ans,ch(s,i,j))
	print(ans)
