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())

l = []
for i in range(10):
    for j in range(i,10):
        l.append([])
        l[-1].append(str(i))
        l[-1].append(str(j))


def solve():
    if len(set(s)) == 1:
        return 0
    val = 0
    for i in range(55):
        l1 = [x for x in s if x in l[i]]
        if len(l1) == 0: continue
        if len(set(l1)) == 1:
            val = max(len(l1),val)
        else:
            newlist = [l1[0]]
            for j in range(1,len(l1)):
                if l1[j] != l1[j-1]: newlist.append(l1[j])
            if len(newlist) % 2 == 1:
                val = max(len(newlist)-1,val)
            else: val = max(len(newlist),val)
    return len(s) - val


for test in range(T):
    s = list(input())
    print(solve())