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())):
    n,k,z=map(int,input().split())
    l=list(map(int,input().split()))
    l1=[]
    l1.append(l[0])
    for i in range(1,k+1):
        l1.append(l[i]+l[i-1])
    temp=l1[0]

    for i in range(1,len(l1)):
        if(l1[i]>temp):
            temp=l1[i]
            ti=i
    ans=l1[0]
    for i in range(1,ti):
        ans+=l[i]
        k-=1
    while(k>1 and z>0):
        ans+=l1[ti]
        k-=2
        z-=1
    for i in range(ti,ti+k):
        ans+=l[i]
    print(ans)"""

for _ in range(int(input())):
    s=str(input())
    l=list(s)
    if(len(list(set(l)))==1):
        print(0)
    else:
        l1=[0]*10
        l2=[-1]*10
        for i in range(len(s)):
            l1[int(s[i])]+=1
            l2[int(s[i])]=1
        ans=min(len(s)-max(l1),len(s)-2)
        l3=[]
        for i in range(10):
            if(l2[i]==1):
                l3.append(i)
        mv=0

        for i in range(len(l3)):

            for j in range(len(l3)):
                f=0
                x1=0
                y1=0
                for k in range(len(s)):
                    if(f==0 and int(s[k])==l3[i]):
                        f=1
                        x1+=1
                    elif(f==1 and int(s[k])==l3[j]):
                        f=0
                        y1+=1
                mv=max(mv,2*(min(x1,y1)))
        print(min(ans,len(s)-mv))
