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.
from sys import stdin, stdout

# 1010
# 12121212
def good_string(s):
    dp = [[0 for _ in range(10)] for _ in range(10)]
    cnt = [0] * 10

    ans1 = 0
    ans2 = 0

    for c in s:
        cv = ord(c) - ord('0')

        cnt[cv] += 1
        ans1 = max(ans1, cnt[cv])

        for i in range(10):
            if cv == i:
                continue

            if dp[cv][i] % 2 == 0:
                dp[cv][i] += 1
            if dp[i][cv] % 2 == 1:
                dp[i][cv] += 1
                ans2 = max(dp[i][cv], ans2)

    return len(s) - max(ans1, ans2)


t = int(stdin.readline())
for _ in range(t):
    s = stdin.readline().strip()
    ans = good_string(s)
    stdout.write(str(ans) + '\n')
