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.

tt = int(input())
while tt!=0:


    n = input()
    d = {}
    for i in range(10):
        d[i]=0
    for i in range(len(n)):
        d[int(n[i])]+=1

    maxi = 0
    for i in d.keys():
        maxi=max(maxi,d[i])
    ans = len(n)-maxi
 
    for i in range(10):
        for j in range(10):

            if i==j:
                continue
            else:
                k=0
                t=0

                while k<len(n):
                    f1, f2 = 0, 0
                    while k<len(n) :
                        if int(n[k])==i:
                            f1=1
                            break
                        k+=1
                    if f1==1:
                        while k < len(n):
                            if int(n[k]) == j:
                                f2 = 1
                                break
                            k += 1
                    if f1==0 or f2==0:
                        break
                    else:
                        t+=2

                ans = min(ans,len(n)-t)


    print(ans)



    tt-=1