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())

for _ in range(t):
    s = str(input())

    if s[-1] + s[:len(s) - 1] == s[1:] + s[0]:
        print(0)
        continue

    s = [x for x in s]

    a = '0123456789'
    d = {x + y: [False, 0] for x in a for y in a}

    for i, x in enumerate(s):
        for y in a:
            z = x + y
            if not d[z][0]:
                d[z][0] = True
                d[z][1] += 1

            z = y + x
            if d[z][0]:
                d[z][0] = False
                d[z][1] += 1

    ans = 0

    for x in a:
        d[x + x][1] //= 2

    for x in d.keys():
        if x[0] == x[1]:
            ans = max(ans, d[x][1])
        else:
            ans = max(ans, int(d[x][1] // 2) * 2)

    print(int(len(s) - ans))
