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.
n=int(input())
for i in range(n):
    s=input()
    l=len(s)
    d=dict()
    for i in "0123456789":
        d[i]=0
    for i in s:
        d[i]=d[i]+1
    d=sorted(d.items(), key=lambda x:x[1],reverse=True)
    maxi=d[0][1]
    arr=d
    for i in range(len(arr)):
        for j in range(len(arr)):
            if(i==j):
                continue
            else:
                e=arr[i][0]
                f=arr[j][0]
                slag=-1
                sount=0
                for q in s:
                    if(q==e):
                        slag=0
                    if(q==f and slag==0):
                        sount=sount+1
                        slag=1
                if(sount*2>=maxi):
                        maxi=sount*2
    print(l-maxi)                
                 