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 math
from collections import deque
from sys import stdin, stdout
from string import ascii_letters
input = stdin.readline
#print = stdout.write
letters = ascii_letters[:26]

def find(p, t, rev):
    if not rev:
        for i in range(len(t) - len(p)):
            if p == t[i:i + len(p)]:
                return i
        return -1
    else:
        for i in range(len(t) - len(p), -1, -1):
            if p == t[i:i + len(p)]:
                return i
        return -1

for _ in range(int(input())):
    x = list(input().strip())

    res = 99999999
    for k in range(10):
        for g in range(10):
            need = 0
            cnt = 0
            for i in x:
                if need:
                    if i == str(g):
                        need = (1 + need) % 2
                        cnt += 1
                else:
                    if i == str(k):
                        need = (1 + need) % 2
                        cnt += 1
            if cnt % 2:
                continue
            res = min(res, len(x) - cnt)
    for k in range(10):
        res = min(res, len(x) - x.count(str(k)))
    res = min(res, len(x) - 2)
    print(res)
