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.
from collections import defaultdict as dd
import math
import sys
input=sys.stdin.readline
def nn():
	return int(input())

def li():
	return list(input())

def mi():
	return map(int, input().split())

def lm():
	return list(map(int, input().split()))

q= nn()

for _ in range(q):
	s = input()

	min_del = dict()
	min_del[(-1,-1)]=0
	for i in range(0,10):
		for j in range(0,10):
			min_del[(i,j)]=0
	for i in range(len(s)-1):
		##min_del[(-1,-1)]=i+1
		for j in range(10):
			min_del[(j,int(s[i]))]=min_del[(int(s[i]), j)] + 1
		##print(min_del)

	m_del = max(min_del.values())
	#print('del:', m_del)
	if (m_del)%2==0:
		print(len(s)-1-m_del)
	else:
		print(min(len(s)-1-m_del+1, len(s)-1-max([min_del[i,i] for i in range(10)])))
