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.
def mergeSort(L,R,a): 
    i = j = k = 0
    while i < len(L) and j < len(R): 
        if L[i] < R[j]: 
            a[k] = 0 
            i+= 1
        else: 
            a[k] = 1 
            j+= 1
        k+= 1
    while i < len(L): 
        a[k] = 0 
        i+= 1
        k+= 1
      
    while j < len(R): 
        a[k] = 1 
        j+= 1
        k+= 1


for _ in range(int(input())):
	s = input()
	arr = [0]*10
	for i in range(len(s)):
		if arr[int(s[i])]==0:
			arr[int(s[i])]=[i]
		else:
			arr[int(s[i])].append(i)
	
	possible = 2
	for i in range(10):
		if arr[i]!=0:
			possible = max(possible,len(arr[i]))
		for j in range(i+1,10):
			if arr[i]!=0 and arr[j]!=0:
				a = [0]*(len(arr[i])+len(arr[j]))
				mergeSort(arr[i],arr[j],a)
				pair = 0
				k = 0
				# print(a)
				while(k<len(a)):
					start = a[k]
					while(k<len(a) and a[k]==start):
						k+=1
					if k == len(a):
						break
					start = a[k]
					while(k<len(a) and a[k]==start):
						k+=1 
					pair+=1
				possible = max(possible,2*pair)
	print(len(s)-possible)

