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; input = sys.stdin.buffer.readline
# sys.setrecursionlimit(10**7)
from collections import defaultdict
con = 10 ** 9 + 7; INF = float("inf")

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

#処理内容
def main():
	T = int(input())
	for _ in range(T):
		s = list(map(int, list(input())))
		N = len(s)
		ans = N
		for i in range(10):
			for j in range(10):
				cnt = 0
				for k in s:
					if cnt % 2 == 0 and k == i:
						cnt += 1
					elif cnt % 2 == 1 and k == j:
						cnt += 1

				if i != j:
					cnt = int(cnt // 2) * 2

				ans = min(ans, N - cnt)

		if N == 1:
			ans = 0
		print(ans)


if __name__ == '__main__':
	main()