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.
"""
                            pppppppppppppppppppp
                         ppppp  ppppppppppppppppppp
                      ppppppp    ppppppppppppppppppppp
                      pppppppp  pppppppppppppppppppppp
                      pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
       ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp
      pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp
     ppppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppppp
    ppppppppppppppppppppppppppppppppppppppppppppppp    pppppppppppppppppppppppp
   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp
  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
                              pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppp  pppppppp
                              ppppppppppppppppppppp    ppppppp
                                 ppppppppppppppppppp  ppppp
                                    pppppppppppppppppppp
"""


import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


for _ in range(int(data())):
    s = data()
    if len(set(s)) == 1 or len(s) <= 2:
        out(0)
        continue
    c = dd(list)
    for i in range(len(s)):
        c[int(s[i])].append(i)
    answer = len(s) - 2
    for i in c.values():
        answer = min(answer, len(s) - len(i))
    for c1 in range(10):
        if not c[c1]:
            continue
        for c2 in range(10):
            if c1 == c2 or not c[c2]:
                continue
            t1 = t2 = cnt = 0
            a, b = c[c1], c[c2]
            par = 0
            while t1 < len(a) and t2 < len(b):
                if par:
                    if a[t1] < b[t2]:
                        t1 += 1
                        par = 0
                        cnt += 1
                    else:
                        t2 += 1
                else:
                    if b[t2] < a[t1]:
                        t2 += 1
                        par = 1
                        cnt += 1
                    else:
                        t1 += 1
            if par and t1 < len(a):
                if a[t1] > b[-1]:
                    cnt += 1
            if not par and t2 < len(b):
                if b[t2] > a[-1]:
                    cnt += 1
            if cnt & 1:
                cnt -= 1
            answer = min(answer, len(s) - cnt)
    out(answer)
