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
import itertools
from collections import defaultdict
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0

lst = [chr(i) for i in range(ord('0'),ord('9')+1)]

def main():
    s = readline().rstrip().decode('utf-8')
    dic1 = defaultdict(int)
    for i in s:
        dic1[i] += 1
    
    ans = max(list(dic1.values()))
    first = 0
    for itr in itertools.combinations(lst,2):
        res = 0
        nec = -1
        for j in s:
            if nec == -1:
                if j == itr[0]:
                    res += 1
                    nec = 1
                    first = 0
                elif j == itr[1]:
                    res += 1
                    nec = 0
                    first = 1
            elif nec == 0:
                if j == itr[nec]:
                    res += 1
                    nec = 1
            elif nec == 1:
                if j == itr[nec]:
                    res += 1
                    nec = 0
        if first != nec: #not parity
            res -= 1
        ans = max(ans,res)
    
    print(len(s) - ans)


if __name__ == "__main__":
    query = int(readline())
    for _ in range(query):
        main()

