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.
import bisect
for _ in range(int(input())):
    st=input()
    dic={}
    le=len(st)
    for i in range(le):
        if(st[i] in dic):
            dic[st[i]].append(i)
        else:
            dic[st[i]]=[i]
    dicle={}
    for i in dic:
        dicle[i]=len(dic[i])
    ans=0
    for i in dic:
        for j in dic:
            if(i!=j):
                ind=dic[i][0]
                ii=0
                while(True):
                    if(ii%2==0):
                        xx=bisect.bisect(dic[j],ind)
                        if(xx==dicle[j]):
                            break
                        else:
                            ind=dic[j][xx]
                    else:
                        xx=bisect.bisect(dic[i],ind)
                        if(xx==dicle[i]):
                            break
                        else:
                            ind=dic[i][xx]
                    ii+=1
                ans=max(ans,(ii+1)//2)
    ans=2*ans
    for i in dicle:
        ans=max(ans,dicle[i])
    print(le-ans)