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,bisect
from collections import defaultdict
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,OrderedDict
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
def neo(): return map(int,input().split())
def Neo(): return list(map(int,input().split()))
for _ in range(int(input())):
	s = input()
	maxi=-10**18
	for i in range(10):
		for j in range(10):
			if(i==j):
				maxa=-10**18
				c2=0
				for z1 in range(len(s)):
					if((ord(s[z1])-48)==i):
					    c2+=1
				maxa=max(c2,maxa)
			else:
				z=0
				c1=0;
				h=len(s)
				while(z<h-1):
				    if((ord(s[z])-48)==i):
					    z+=1
					    while(z!=h and (ord(s[z])-48)!=j):
						    z+=1
					    if(z!=h):
						    c1+=1
						    z+=1
				    else:
					    z+=1
				maxi=max(maxi,2*c1)
			maxi=max(maxa,maxi);
	print(len(s)-maxi)
