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
from collections import deque
input = sys.stdin.readline
t = int(input())
sub = []
for i in range(10):
    for j in range(10):
        sub.append((i,j))
for _ in range(t):
    s = list(input())
    s = [int(s[i]) for i in range(len(s)-1)]
    ans = float('inf')
    if len(s)<=2:
        print(0); continue
    for a,b in sub:
        new = []
        for i in s:
            if i==a or i==b: new.append(i)
        n = len(new)
        c = len(s)-n
        new = deque(new)
        d = (a,b)
        nn = []
        while new:
            curr = new.popleft()
            if nn:
                if nn[-1]!=curr:
                    nn.append(curr)
            else:
                nn.append(curr)
        c+=n-len(nn)
        if len(nn)%2:
            c+=1
        ans = min(ans, c)
    for i in range(10):
        ans = min(ans, len(s)-s.count(i))
    print(ans)
