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 get_a(even,odd,a):
    # print(even,odd)
    flag=0
    b=[]
    for i in range(0,len(a)):
        if(flag==0 and a[i]==even):
            b.append(even)
            flag=1
        elif(flag==1 and a[i]==odd):
            b.append(odd)
            flag=0
    # print(b,len(b))
    return b


def process(n,a):
    a=list(a)
    m=0
    for i in range(0,10):
        for j in range(0,10):
            a1=get_a(str(i),str(j),a)
            if(len(a1)%2==0 or i==j):
                m=max(m,len(a1))
    return len(a)-m



t=int(input())
for i in range(0,t):
    # n=int(input())
    a=input()
    # print(len(a))
    print(process(len(a),a))
