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 math
from functools import reduce
import bisect


def getN():
    return int(input())


def getNM():
    return map(int, input().split())


def getList():
    return list(map(int, input().split()))


def input():
    return sys.stdin.readline().rstrip()


def index(a, x):
    i = bisect.bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    return False


#############
# MAIN CODE #
#############

for _ in range(int(input())):
    s = [int(x) for x in input()]
    ans = 0

    for i in range(10):
        for j in range(10):
            t = 0
            k = i

            if i ==  j:
                continue

            for p in s:
                if p == k:
                    if k == j:
                        t += 1
                        k = i
                    else:
                        k = j
            ans = max(ans, t)


    nums = [0] * 10

    for i in s:
        nums[i] += 1

    ans *= 2
    ans = max(ans, max(nums))

    print(max(len(s) - ans, 0))
