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.
def solve (a,b,s):
    res=0;si=1

    for i in range(0,len(s)):
        if (ord(s[i])-ord('0')==a and si==1) :
            res+=1
            si=-si
        elif (ord(s[i])-ord('0')==b and si==-1) :
            res+=1
            si=-si

    if (res&1 and res>1 and a!=b) : res-=1

    return res

tc =int (input())

while (tc):
    s=input()

    max1=0
    for i in range (0,10):
        for j in range (0,10):
            max1=max(max1,solve(i,j,s))

    print (len(s)-max1)
    tc-=1
