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 collections import Counter
d = [str(i)+str(j) for i in range(10) for j in range(10)]
cases = int(input())
for t in range(cases):
    n = input()
    cc = Counter(n)
    mv = max(cc.values())
    for i in d:
        s = 0
        a,b = i[0],i[1]
        f1 = n.find(a)
        f2 = n.find(b,f1+1)
        while True:
            s += 2
            f1 = n.find(a,f2+1)
            f2 = n.find(b,f1+1)
            if f1==-1 or f2==-1:
                break
        mv = max(mv,s)
    print(len(n)-mv)
