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 everythingfine(s,n):
    for i in range(n):
        if s[i]!=s[(i+2)%n]:
            return False
    return True
for _ in range(int(input())):
    s=input()
    n = len(s)
    maxi = 0
    temp = list("0123456789")
    d=dict()
    for i in temp:
        d[i]=s.count(i)
        maxi = max(maxi,d[i])
    
    if everythingfine(s,n):
        print("0")
        continue
    
    #trying for odd solution
    ans = min(n-2,n-maxi)

    #trying for even solution
    for i in range(10):
        for j in range(10):
            if i==j or d[temp[i]]==0 or d[temp[j]]==0:
                continue
            tempans = 0
            diff = 0
            for x in s:
                if x==temp[i] and diff==1:
                    tempans+=1
                elif x==temp[j] and diff==0:
                    tempans+=1
                elif x==temp[i] and diff==0:
                    diff+=1
                elif x==temp[j] and diff==1:
                    diff-=1
                else:
                    tempans+=1
            if diff==0:
                ans = min(ans,tempans)
    print(ans)
    