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.
for t in range(int(input())):
    s=input()
    i=0
    n=len(s)
    if (n<=2):
        print(0)
        continue
    ans=2
    freq={}
    maxF=1
    for c in s:
        if c not in freq:
            freq[c]=1
        else:
            freq[c]+=1
            maxF=max(maxF,freq[c])
    ans=max(ans,maxF)
    for i in range(0,10):
        for j in range(0,10):
            if (i!=j):
                x=str(i)
                y=str(j)
                val=0
                target=x
                for k in range(n):
                    if (s[k]==target):
                        val+=1
                        if target==x:
                            target=y
                        else: target=x
                if (val%2!=0): val-=1
                ans=max(ans,val)
    print(n-ans)
