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 sys import stdin, exit, setrecursionlimit
from collections import deque

input = stdin.readline
lmi = lambda: list(map(int, input().split()))
mi = lambda: map(int, input().split())
si = lambda: input().strip('\n')
ssi = lambda: input().strip('\n').split()

for _ in range(int(input())):
    s = list(map(int, list(si())))
    freq = [0 for i in range(10)]
    alt = [[0 for j in range(10)] for i in range(10)]
    vis = [[True for j in range(10)] for i in range(10)]
    for i in s:
        freq[i] += 1
        for j in range(10):
            if j != i:
                vis[i][j] = False
            # vis[j][i] = False
        for j in range(10):
            if not vis[j][i]:
                vis[j][i] = True
                alt[j][i] += 1
    mx = 0
    for i in alt:
        mx = max(max(i), mx)
    freq.sort()
    ans = freq[-1]
    print(len(s) - max(mx*2, ans))