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.
import math
import sys
import bisect
#input=sys.stdin.readline
t=int(input())
#t=1
for _ in range(t):
    #n,k,z=map(int,input().split())
    #l=list(map(int,input().split()))
    s=input()
    d={}
    for i in range(10):
        d[i]=[]
    for i in range(len(s)):
        d[int(s[i])].append(i)
    maxi=0
    for i in range(10):
        maxi=max(maxi,len(d[i]))
    for i in range(10):
        for j in range(10):
            if i!=j and len(d[i])>0 and len(d[j])>0:
                cnt=0
                i1=-1
                while 1:
                    i1=bisect.bisect(d[i],i1)
                    
                    if i1>=len(d[i]):
                        break
                    i1=d[i][i1]
                    #print(1,i,i1)
                    i1=bisect.bisect(d[j],i1)
                    if i1>=len(d[j]):
                        break
                    i1=d[j][i1]
                    #print(2,j,i1)
                    cnt+=2
                maxi=max(maxi,cnt)
    print(len(s)-maxi)
    

            
        