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.
from collections import defaultdict
def solve():
    s = raw_input().strip()
    n = len(s)
    d = defaultdict(int)
    for c in s:
        d[c] += 1
    ans = n - max(d.values())
    t = d.keys()
    l = len(t)
    a = ["".join(t).translate(None, t[i] + t[j]) for i in xrange(l) for j in xrange(i + 1, l)]
    e = [s.translate(None, p) for p in a]
    for c in e:
        l = len(c)
        x = sum(c[i+1] != c[i] for i in xrange(l-1)) + 1
        x = n - x / 2 * 2
        if ans > x:
            ans = x
    print ans

T = int(raw_input())
for _ in xrange(T):
    solve()
