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()
    n = len(s)
    cnt = [0] * 10
    ind = [[] for i in range(10)]
    for i in range(n):
        cnt[int(s[i])] += 1
        ind[int(s[i])] += [i]
    ans = n - max(cnt)
    for i in range(10):
        for j in range(i + 1, 10):
            l = 0
            r = 0
            cur = []
            a = 0
            fl = -1
            for k in range(cnt[i] + cnt[j]):
                if l == cnt[i]:
                    cur.append(ind[j][r])
                    r += 1
                    if fl == 0:
                        a += 1
                    fl = 1
                elif r == cnt[j]:
                    cur.append(ind[i][l])
                    if fl == 1:
                        a += 1
                    fl = 0
                    l += 1
                else:
                    if ind[i][l] < ind[j][r]:
                        cur.append(ind[i][l])
                        l += 1
                        if fl == 1:
                            a += 1
                        fl = 0
                    else:
                        cur.append(ind[j][r])
                        r += 1
                        if fl == 0:
                            a += 1
                        fl = 1
            a += 1
            ans = min(ans, n - a + (a% 2))
    print(ans)