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.
n = int(input())

for i in range(n):
    m = 0
    arr1 = str(input())
    for j in range(0, 10):
        for k in range(0, 10):
            if j < k:
                coun = 0
                y = 0
                for t in range(len(arr1)):
                    if coun == 0 and arr1[t] == str(j):
                        coun += 1
                        y = str(j)
                    elif coun == 0 and arr1[t] == str(k):
                        coun += 1
                        y = str(k)
                    elif arr1[t] == str(j) and y == str(k):
                        coun += 1
                        y = str(j)
                    elif arr1[t] == str(k) and y == str(j):
                        coun += 1
                        y = str(k)
                if coun % 2 == 0 and coun > m:
                    m = coun
                elif coun % 2 == 1 and coun - 1 > m:
                    m = coun - 1
    for j in range(10):
        if arr1.count(str(j)) > m:
            m = arr1.count(str(j))
    print(len(arr1) - m)
