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
from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right
input = sys.stdin.readline
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().split()))
v=[[str(i),str(j)] for i in range(10) for j in range(10) if i!=j]
for _ in range(inp()):
    s=insr()
    n=len(s)
    c=Counter(s)
    mx=max(c.values())
    cnt=0
    for a,b in v:
        temp=0
        cntb=0
        for i in range(n-1,-1,-1):
            if s[i]==b:
                cntb=1
            if s[i]==a:
                if cntb:
                    temp+=2
                    cntb=0
            if temp%2==1 and a!=b:
                temp-=1
        cnt=max(temp,cnt)
    print(n-max(cnt,mx))