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())):
    s = input()
    dp = [[0 for i in range(10)] for j in range(10)]
    mx = 0
    for i in range(10):
        sl = str(i)
        for j in range(10):
            nl = str(j)
            fi , ei = -1 , -1
            used = 0
            for k in range(len(s)):
                if i == j and s[k] == sl:
                    dp[i][i] += 1
                    continue
                elif i == j:
                    continue
                else:
                    if s[k] == sl:
                        fi = k
                        used = 0
                    elif s[k] == nl:
                        ei = k
                    if not used and ei > fi and fi >= 0:
                        dp[i][j] += 1
                        used = 1
                    #print( i , j , fi , ei)
            if i != j:
                mx = max(mx , dp[i][j])
    #print(dp)
    mx = max(mx * 2 , max(dp[i][i] for i in range(10)))
    print(len(s) - mx)