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,sys
from sys import stdin,stdout
from collections import Counter, defaultdict, deque
input = stdin.readline
I = lambda:int(input())
li = lambda:list(map(int,input().split()))

def solve():
    a=input().strip()
    n=len(a)
    m=n
    for i in range(0,10):
        for j in range(0,10):
            if(i==j):
                continue
            c=0 
            p=[str(i),str(j)]
            k=0 
            t=0
            while(k<n):
                if(a[k]==p[t]):
                    t+=1
                if(t==2):
                    t=0
                    c+=2
                k+=1
            m=min(m,n-c)
    c=Counter(a)
    for i in c:
        m=min(m,n-c[i])
    print(m)

    
    
for _ in range(I()):
    solve()