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
INP = lambda: sys.stdin.readline().strip()
INT = lambda: int(INP())
MAP = lambda: map(int, INP().split())
ARR = lambda: [int(i) for i in INP()]
def JOIN(arr, x=' '): return x.join([str(i) for i in arr])
def EXIT(x='NO'): print(x); exit()

def do(f,arr,a,b):
    xx = 0
    for x in arr:
        if x==f:
            if x==a: f=b
            else: f=a
            xx += 1
    return (xx//2)*2

def solve():
    arr = [int(x) for x in list(INP())]
    c = {x: 0 for x in arr}
    for x in arr: c[x] += 1
    cs = sorted(list(c.items()), key=lambda x: x[1], reverse=True)
    if len(cs)<2: return 0
    ans = cs[0][1]
    l = len(cs)
    for i in range(l):
        for j in range(i+1,l):
            ans = max(ans, do(cs[i][0], arr, cs[i][0], cs[j][0]))
            ans = max(ans, do(cs[j][0], arr, cs[i][0], cs[j][0]))
    return len(arr) - ans


for _ in range(INT()): print(solve())