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.
from sys  import stdin,stdout

import bisect

import math

def st():
    return list(stdin.readline().strip())

def inp():
    return int(stdin.readline())

def li():
    return list(map(int,stdin.readline().split()))

def mp():
    return map(int,stdin.readline().split())

def pr(n):
    stdout.write(str(n)+"\n")

def soe(limit):
    l=[1]*(limit+1)
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0

    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime

def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
                
def gcd(a,b):
    while b:
        a,b=b,a%b
    return a

def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
        a*=a
        n=n>>1
    return r

    
def solve():
    s=input()
    n=len(s)
    d={}
    for i in s:
        d[i]=d.get(i,0)+1
    mi=float('-inf')
    for i in d:
        if d[i]>mi:
            mi=d[i]
    for i in range(10):
        for j in range(10):
            c=0
            for k in s:
                if c%2:
                    if k==str(j):
                        c+=1
                else:
                    if k==str(i):
                        c+=1
            if c&1:
                c-=1
            mi=max(c,mi)
    pr(n-mi)
    
for _ in range(inp()):
    solve()
    
