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(o, p):
    c = c2 = c1 = c0 = 0
    for i in range (n):
        if s[i] == p:
            c1 = 1
        if c1 == 1:
            if s[i] == o:
                c += 1
                c1 = 0
    for i in range (n):
        if s[i] == o:
            c2 = 1
        if c2 == 1:
            if s[i] == p:
                c0 += 1
                c2 = 0
    return max(c, c0)
for _ in range (int(input())):
    s = input()
    l = []
    for i in range (10):
        l.append([i, 0])
    for i in s:
        l[int(i)][1] += 1
    l.sort(key = lambda x : x[1], reverse = True)
    k = l[1][1]
    n = len(s)
    su = 0
    for i in range (10):
        for j in range (i + 1, 10):
            su = max(su, fun(str(l[i][0]), str(l[j][0])))
    #print(su)
    if l[1][1] == 0:
        print(0)
    else:
        print(min(n - 2 * su, n - l[0][1]))