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.
import sys
readline = sys.stdin.readline

def solve():
    S = readline().rstrip()
    N = len(S)

    odds = [S.count(str(i)) for i in range(10)]
    odds_max = max(odds)

    ans = N - max(odds)

    for i in range(10):
        if odds[i] * 2 <= odds_max:
            continue
        for j in range(i + 1, 10):
            if odds[j] * 2 <= odds_max:
                continue
            
            count = 0
            last = None
            _i, _j = str(i), str(j)
            for v, c in enumerate(S):
                if c != last:
                    if c == _i:
                        last = _i
                        count += 1
                    elif c == _j:
                        last = _j
                        count += 1
            ans = min(ans, N - count // 2 * 2)
    
    print(ans)

T = int(readline())
for t in range(T):
    solve()