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
range = xrange
input = raw_input

t = int(input())
for _ in range(t):
    A = [ord(c) - ord('0') for c in input()]
    n = len(A)

    nesta = [n] * 10
    B = [list(nesta)]
    B.append(list(nesta))
    for i in reversed(range(n)):
        nesta[A[i]] = i
        B.append(list(nesta))
    B.reverse()
    
    counter = [0]*10
    for a in A:
        counter[a] += 1
    mini = n - max(counter)
    for i in range(100):
        b, a = divmod(i, 10)
        count = 0
        j = 0
        while True:
            j = B[j][a] + 1
            j = B[j][b] + 1
            if j > n:
                break
            count += 1
        mini = min(mini, n - 2 * count)
    print mini

