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.
def fun(S, s):
    N = len(S)
    i = 0
    flip = 1
    cun= 0
    while(i<N):
        if(flip):
            if(S[i] == s[0]):
                i+=1
                flip  = 1 - flip
                continue
            
        else:
            if(S[i] == s[1]):
                flip = 1 - flip
                cun += 1
                i+=1
                continue
        i+=1
            
    return cun


for _ in range(int(input())):
    S = input()
    N = len(S)
    #Always
    if(N<=2):
        print(0)
        continue
    Always = N - 2
    #Solo
    Single = 0
    for i in range(10):
        s = str(i)
        count = S.count(s)
        if(count != -1):
            Single = max(Single, count)
    #Double barrel
    Double = 0

    for i in range(0, 100):
        if(i<10):
            s = "0" + str(i)
        else:
            s = str(i)
        count = fun(S, s)
        Double = max(Double, count)
    X = min(Always, N-Single, N-2*Double)
    print(X)
    

