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 collections import defaultdict
for _ in range(int(input())):
    s=input()
    d=defaultdict(int)
    st=set()
    for i in s:
        st.add(i)
    if(len(st)==1):
        print(0)
        continue
    digit=['00','01','02','03','04','05','06','07','08','09']

    for i in range(10,100):
        digit.append(str(i))

    mx=0
    for i in digit:
        ct=0
        sec=False
        for j in s:
            if(sec==False and j==i[0]):
                ct+=1
                sec=True
                continue
            if(sec==True and j==i[1]):
                sec=False
                ct+=1
        if(ct%2==1 and i[0]!=i[1]):
            ct-=1
        mx=max(ct,mx)
    print(len(s)-mx)
