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.
t=int(input())
for i in range(t):
    s=str(input())
    l=len(s)
    a0=a1=a2=a3=a4=a5=a6=a7=a8=a9=0
    a=l-2
    for i in s:
        if i=='0':
            a0=a0+1
        elif i=='1':
            a1=a1+1
        elif i=='2':
            a2=a2+1
        elif i=='3':
            a3=a3+1
        elif i=='4':
            a4=a4+1
        elif i=='5':
            a5=a5+1
        elif i=='6':
            a6=a6+1
        elif i=='7':
            a7=a7+1
        elif i=='8':
            a8=a8+1
        else:
            a9=a9+1
    bm=max(a0, a1, a2, a3, a4, a5, a6, a7, a8, a9)
    b=l-bm
    m=0
    for i in range(10):
        for j in range(10):
            c=0
            if i!=j:
                for x in s:
                    y=int(x)
                    if c%2==0 and y==i:
                        c=c+1
                    elif c%2==1 and y==j:
                        c=c+1
                if c>m:
                    m=c
    m=m//2
    m=m*2
    d=l-m
    print(max(min(a, b, d), 0))