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.
t = int(input())
while t>0:
	a = list(str(input()))
	d = {}
	d['0'] = []
	d['1'] = []
	d['2'] = []
	d['3'] = []
	d['4'] = []
	d['5'] = []
	d['6'] = []
	d['7'] = []
	d['8'] = []
	d['9'] = []
	for i in range(len(a)):
		d[a[i]].append(i)
	co = 0
	g = d.keys()
	for i in d.keys():
		co = max(co,len(d[i]))
	# print(d,co)
	for i in range(9):
		a1 = d[str(i)]
		if len(a1) == 0:
			continue
		for j in range(i+1,10):
			a1 = d[str(i)]
			a2 = d[str(j)]
			if len(a2) == 0:
				continue
			count = 0
			if a1[0]>a2[0]:
				# print(a1,a2)
				a1d = []
				for k in range(len(a1)):
					a1d.append([a1[k],1])
				# print(a1)
				a2d = []
				for k in range(len(a2)):
					a2d.append([a2[k],0])
				# print(a2)
				c = list(a1d)
				# print(c)
				c.extend(a2d)
				# print(c)
				c.sort(key =lambda x:x[0])
				ini = 0
				for k in range(len(a1d)+len(a2d)):
					# print(c,k,a1,a2)
					if ini == c[k][1]:
						count += 1
						if ini == 0:
							ini = 1
						else:
							ini = 0
				if count%2 != 0:
					count = count-1
			else:
				# print(a1,a2)
				a1d = []
				for k in range(len(a1)):
					a1d.append([a1[k],1])
				# print(a1)
				a2d = []
				for k in range(len(a2)):
					a2d.append([a2[k],0])
				# print(a2)
				c = list(a1d)
				# print(c)
				c.extend(a2d)
				# print(c)
				c.sort(key =lambda x:x[0])
				ini = 1
				for k in range(len(a1d)+len(a2d)):
					# print(c,k,a1,a2)
					if ini == c[k][1]:
						count += 1
						if ini == 0:
							ini = 1
						else:
							ini = 0
				if count%2 != 0:
					count = count-1
			if count>co:
				co = count
	print(len(a)-co)
	t = t-1