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

read = sys.stdin.buffer.read
# input = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
t = int(input())
for case in range(t):
    s = input()
    ans = 0
    for i in range(10):
        for j in range(10):
            i_bit = 0  # 0ならi かどうかを判定する
            ij_len = 0
            i_len = 0
            for k in s:
                if i_bit == 0:
                    if i == int(k):
                        i_bit = 1
                        ij_len += 1
                else:
                    if j == int(k):
                        i_bit = 0
                        ij_len += 1
                if i == int(k):
                    i_len += 1
            # print(i, j, ij_len, i_len, ans)
            ans = max(ans, i_len, (ij_len // 2) * 2)
    print(min(len(s) - 2, len(s) - ans))

