Chef Al Gorithm was reading a book about climate and oceans when he encountered the word “glaciological”. He thought it was quite curious, because it has the following interesting property: For every two letters in the word, if the first appears x times and the second appears y times, then |x - y| ≤ 1.
Chef Al was happy about this and called such words 1-good words. He also generalized the concept: He said a word was K-good if for every two letters in the word, if the first appears x times and the second appears y times, then |x - y| ≤ K.
Now, the Chef likes K-good words a lot and so was wondering: Given some word w, how many letters does he have to remove to make it K-good?

-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
Each test case consists of a single line containing two things: a word w and an integer K, separated by a space.

-----Output-----
For each test case, output a single line containing a single integer: the minimum number of letters he has to remove to make the word K-good.

-----Constraints-----
- 1 ≤ T ≤ 30
- 1 ≤ |w| ≤ 105
- 0 ≤ K ≤ 105
- w contains only lowercase English letters.

-----Example-----
Input:
4
glaciological 1
teammate 0
possessions 3
defenselessness 3

Output:
0
0
1
2

-----Explanation-----
Example case 1. The word “glaciological” is already 1-good, so the Chef doesn't have to remove any letter.
Example case 2. Similarly, “teammate” is already 0-good.
Example case 3. The word “possessions” is 4-good. To make it 3-good, the Chef can remove the last s to make “possession”.
Example case 4. The word “defenselessness” is 4-good. To make it 3-good, Chef Al can remove an s and an e to make, for example, “defenslesness”. Note that the word doesn't have to be a valid English word.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '\nt = int(input())\nwhile t:\n w,k = input().split()\n k = int(k)\n char = {}\n for c in w:\n  if c not in list(char.keys()):\n   char[c] = 1\n  else:\n   char[c] += 1\n counts = list(char.values())\n counts.sort()\n #counts = counts[::-1]\n l = len(counts)\n ans = len(w)+1\n for i in range(l):\n  temp = 0\n  for j in range(l):\n   if counts[j]<counts[i]:\n    temp += counts[j]\n   elif counts[j]>counts[i]+k:\n    temp += counts[j]-counts[i]-k\n  ans = min(ans,temp)\n print(ans)\n t -= 1\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13393