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
input = sys.stdin.readline
import math

############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().strip().split()))



def from_file(f):
    return f.readline


def count(l, a, b):
    res = 0
    expect = a
    for c in l:
        if int(c) != expect:
            res += 1
        else:
            expect = b if expect == a else a

    left = len(l) - res
    if left % 2 == 1 and a != b:
        res += 1
    return res


def solve(l):
    m = len(l)
    for a in range(10):
        for b in range(10):
            m = min(m, count(l, a, b))
    # m = count(l, 2,5)
    return m



# with open('3.txt') as f:
#     input = from_file(f)
t = inp()
for _ in range(t):
    l = input().strip()
    print(solve(l))