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 __ in range(int(input())):
    ar = list(map(int, input()))
    kek = [[] for i in range(10)]
    i = 0
    for elem in ar:
        kek[elem].append(i)
        i += 1

    ans = 0
    for i in range(10):
        z = 0
        fifi = [0] * 10
        while z < len(kek[i]) - 1:
            lol = set()
            for j in range(kek[i][z] + 1, kek[i][z + 1]):
                lol.add(ar[j])
            for elem in lol:
                fifi[elem] += 1
            z += 1
        for j in range(10):
            flag = 0
            if kek[i]:
                for z in range(kek[i][0]):
                    if ar[z] == j:
                        flag = 2
            ans = max(ans, fifi[j] * 2 + flag, len(kek[i]))
    print(len(ar) - ans)