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.
def solve (stt):
    leni = len(stt)
    ls = list(stt)
    maxi =-1
    for i in range(0,10):
        for j in range(0,10):
            cr = str(i)+str(j)
            newcr = cr*(leni//2)
            if(leni%2==1):
                newcr=newcr+str(i)
            lss = list(newcr)
            ans = 0
            k1 =0
            k2 =0
            while(k1<leni):
                if(ls[k1]==lss[k2]):
                    k1+=1
                    k2+=1
                    ans+=1
                else:
                    k1+=1
            if(i!=j and ans%2==1):
                ans = -1
            maxi = max(maxi , ans)
    return leni-maxi
    
for i in range(int(input())):
    s = input()
    print(solve(s))