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 numOfChars(s,c):
    ans = 0
    for k in s:
        if k == c:
            ans += 1
    return ans
def calcMaxSubseq(s,charOne,charTwo):
    maxL = 0
    counter = 0
    for c in s:
        if counter == 0 and c == charOne:
            counter = 1
            maxL += 1
        elif counter == 1 and c == charTwo:
            counter = 0
            maxL += 1
    if counter == 1:
        return maxL - 1
    else:
        return maxL
t = int(input())
for i in range(t):
    s = input()
    l = len(s)
    maxL = 0
    print
    for j in range(10):
        maxL = max(maxL,numOfChars(s,str(j)))
        for k in range(j+1,10):
            maxL = max(maxL,calcMaxSubseq(s,str(j),str(k)))
            maxL = max(maxL,calcMaxSubseq(s,str(k),str(j)))
    print(l - maxL)