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.
def hhh(s, u, v):
    c = b = 0
    for i in range(len(s)):
        if b == 0:
            if s[i] == u:
                x = u
                y = v
                b = 1
            elif s[i] == v:
                x = v
                y = u
                b = 1
        elif b == 1:
            if s[i] == y:
                c += 2
                b = 2
        else:
            if s[i] == x:
                b = 1
    return c


for t in range(int(input())):
    s = input()
    a = []
    for i in range(10):
        a.append(s.count(str(i)))
    m = max(a)
    for i in range(10):
        for j in range(i + 1, 10):
            n = hhh(s, str(i), str(j))
            if n > m:
                m = n
    print(len(s)-m)
