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 collections import defaultdict
for _ in range(int(input())):
#     n,k,z = list(map(int, input().split()))
    x = input()
    n = len(x)
    a = []
    for i in x:
        a.append(int(i))
#     a = list(map(int, x.split('')))
    c = [0]*10
    d = [[[0, -1] for _ in range(10)]for _ in range(10)]
    v = [0]*10
    c[a[0]] = 1
    count = 2
    v[a[0]] = 1
#     print(d[9],v,c)
    for i in range(1, n):
        no = a[i]
        c[no]+=1
        if(v[no] == 0):
            for j in range(10):
                if(v[j] != 0):
                    d[j][no][0] = 2
                    d[j][no][1] = no
            v[no] = count
            count+=1
        else:
#             print("a")
            for j in range(10):
                if(d[no][j][1] == j):
                    d[no][j][0]+=1
                    d[no][j][1] = no
            for j in range(10):
                if(d[j][no][1] == j):
                    d[j][no][0]+=1
                    d[j][no][1] = no
#         print(d[0],d[1])
    cm = max(c)
    m = 2
    for i in range(10):
        for j in range(10):
            if(i==j):
                continue
            m = max(d[i][j][0], m)
    print(n - max(cm,m//2*2))