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):
    lin=input()
    a={'0':0,'1':0,'2':0,'3':0,'4':0,'5':0,'6':0,'7':0,'8':0,'9':0}
    b={'0':-1,'1':-1,'2':-1,'3':-1,'4':-1,'5':-1,'6':-1,'7':-1,'8':-1,'9':-1}
    bp={'0':-1,'1':-1,'2':-1,'3':-1,'4':-1,'5':-1,'6':-1,'7':-1,'8':-1,'9':-1}
    enct=[[0 for k in range(10)] for l in range(10)]
    for j in range(len(lin)):
        a[lin[j]]+=1
        bp[lin[j]]=b[lin[j]]
        b[lin[j]]=j
        for k in range(10):
            if b[str(k)]!=-1 and bp[lin[j]]<b[str(k)] and b[lin[j]]>b[str(k)]:
                enct[int(lin[j])][k]+=1
    max1=0
    max2=0
    for j in range(10) :
        max1=max(a[str(j)],max1)
        max2=max(max(enct[j]),max2)
    print(min(len(lin)-(2*max2),len(lin)-max1))