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 i in range(int(input())):
    a=input()
    n=len(a)
    d=n
    for j in range(10):
        e=[0]*(9-j)
        c=0
        g=0
        h=[0]*(9-j)
        for k in range(n):
            y=int(a[k])
            if y==j:
                if c>0:
                    for l in range(9-j):
                        e[l]+=f[l]
                f=[0]*(9-j)
                c+=1
            elif c>0 and y>j:
                z=y-j-1
                if f[z]==0:
                    f[z]=1
            elif c==0 and y>j:
                z=y-j-1
                if h[z]==0:
                    h[z]=1
        if c>0:
            for l in range(9-j):
                e[l]+=f[l]
        c0=c
        for k in range(9-j):
            if h[k]==1 and c0>0 and f[k]==0:
                e[k]+=1
            c=max(c,2*e[k])
        d=min(d,n-c)
    print(d)