Alice and Bob created $N$ and $M$ recipes, respectively ($N, M \ge 1$), and submitted them to Chef for evaluation. Each recipe is represented by a string containing only lowercase English letters. Let's denote Alice's recipes by $A_1, A_2, \ldots, A_N$ and Bob's recipes by $B_1, B_2, \ldots, B_M$.
Accidentally, Chef mixed up those recipes ― now, he has $L = N+M$ recipes in a sequence $S_1, S_2, \ldots, S_L$. Thankfully, the recipes created by Alice and Bob are distinguishable from each other. It is well-known that for each recipe $s$ created by Alice, the following property holds, and for each recipe created by Bob, it does not hold: For each $1 \le l < r \le |s|$, the substring $s_l, s_{l+1}, \ldots, s_r$ contains at least as many vowels as consonants. The letters 'a', 'e', 'i', 'o', 'u' are vowels, while the other letters are consonants.
The score of a candidate who made $K$ recipes is calculated as the product of $\frac{x_c}{fx_c^K}$ for all letters $c$ that occur in at least one of these recipes; here, $x_c$ is the number of recipes which contain the letter $c$ and $fx_c$ is the total number of occurrences of this letter in all $K$ recipes.
Let's denote the scores of Alice and Bob by $sc_A$ and $sc_B$ respectively. Chef wants to know their ratio $sc_A/sc_B$. We know that Chef is a legendary cook, but he is not very good at calculating, so he is asking you to find that number.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains a single integer $L$.
- $L$ lines follow. For each valid $i$, the $i$-th of these lines contains a single string $S_i$.

-----Output-----
For each test case, if the ratio of scores exceeds $10^7$, print a single line containing the string "Infinity" (without quotes); otherwise, print a single line containing one real number $sc_A/sc_B$.
Your answer will be considered correct if its absolute or relative error does not exceed $10^{-6}$. It is guaranteed that $sc_A/sc_B$ does not lie in the range $10^7 \pm 10$.

-----Constraints-----
- $1 \le T \le 10^5$
- $2 \le L \le 10^5$
- $2 \le |S_i| \le 10^5$ for each valid $i$
- for each valid $i$, $S_i$ contains only lowercase English letters
- the sum of $|S_1| + |S_2| + \ldots + |S_L|$ over all test cases does not exceed $10^7$

-----Subtasks-----
Subtask #1 (25 points):
- $L \le 10$
- $|S_i| \le 10$ for each valid $i$
Subtask #2 (75 points): original constraints

-----Example Input-----
2
4
aba
abc
bab
aac
3
aba
baab
abc

-----Example Output-----
1.1250000
0.0277778

-----Explanation-----
Example case 1: The recipes "aba" and "aac" are created by Alice, while the recipes "abc" and "bab" are created by Bob. The scores are:
- $sc_A = \frac{x_a}{fx_a^N} \cdot \frac{x_b}{fx_b^N} \cdot \frac{x_c}{fx_c^N} = \frac{2}{4^2} \cdot \frac{1}{1^2} \cdot \frac{1}{1^2} = \frac{1}{8}$
- $sc_B = \frac{x_a}{fx_a^M} \cdot \frac{x_b}{fx_b^M} \cdot \frac{x_c}{fx_c^M} = \frac{2}{2^2} \cdot \frac{2}{3^2} \cdot \frac{1}{1^2} = \frac{1}{9}$
- $\frac{sc_A}{sc_B} = \frac{1/8}{1/9} = 1.125$

```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 = '# v = ["a","e","i","o","u"]\n# for _ in range(int(input())):\n#   n = int(input())\n#   a,b = [],[]\n#   for i in range(n):\n#       s = input()\n#       isa = True\n#       for j in range(1,len(s) - 1):\n#           if(s[j] in v):\n#               if(s[j - 1] not in v and s[j + 1] not in v):\n#                   isa = False\n#           else:\n#               if(s[j - 1] not in v or s[j + 1] not in v):\n#                   isa = False\n#           if(not isa):\n#               break\n#       if(s[0] not in v and s[1] not in v):\n#           isa = False\n#       if(s[-1] not in v and s[-2] not in v):\n#           isa = False\n#       if(isa):\n#           a.append(s)\n#       else:\n#           b.append(s)\n#   dicta,dictb = {},{}\n#   for i in a:\n#       freq = {}\n#       for j in i:\n#           if(j in freq):\n#               freq[j] += 1\n#           else:\n#               freq[j] = 1\n#       for j in freq:\n#           if(j not in dicta):\n#               dicta[j] = (1,freq[j])\n#           else:\n#               dicta[j] = (dicta[j][0] + 1,dicta[j][1] + freq[j])\n#   for i in b:\n#       freq = {}\n#       for j in i:\n#           if(j in freq):\n#               freq[j] += 1\n#           else:\n#               freq[j] = 1\n#       for j in freq:\n#           if(j not in dictb):\n#               dictb[j] = (1,freq[j])\n#           else:\n#               dictb[j] = (dictb[j][0] + 1,dictb[j][1] + freq[j])\n#   ans = 1\n#   for i in dicta:\n#       ans *= dicta[i][0]\n#   for i in dictb:\n#       ans /= dictb[i][0]\n#   x,y = 1,1\n#   for i in dictb:\n#       x *= dictb[i][1]\n#   for i in dicta:\n#       y *= dicta[i][1]\n#   alice,bob = len(a),len(b)\n#   for i in range(bob):\n#       while(alice > 0 and ans > 10**7):\n#           ans /= y\n#           alice -= 1\n#       ans *= x\n#       if(ans > 10**7 and alice == 0):\n#           break\n#   while(alice > 0):\n#       ans /= y\n#       if(ans < 1 and alice > 100):\n#           ans = 0\n#           break\n#       alice -= 1\n#   if(ans > 10**7):\n#       print("Infinity")\n#   else:\n#       print(ans)\n# #partailly correct [75 pts]\n#sys.stdin.readline() and sys.stdout.write() are faster I/O methods than input() \xa0\xa0\xa0\xa0and print()\nfrom sys import stdin\nz=[\'a\',\'i\',\'e\',\'o\',\'u\']\nt=int(stdin.readline())\nwhile(t>0):\n t-=1\n n=int(stdin.readline())\n alice=[]\n bob=[]\n for j in range(n):\n  s=str(stdin.readline().strip("\\n"))\n  # print(s)\n  isalice=True\n  for i in range(1,len(s)-1):\n   if(s[i] in z):\n    if(s[i-1] not in z and s[i+1] not in z):\n     isalice=False\n   else:\n    if(s[i-1] not in z or s[i+1] not in z):\n     isalice=False\n   if(not isalice):\n    break\n  if(s[0] not in z and s[1] not in z):\n   isalice=False\n  if(s[-1] not in z and s[-2] not in z):\n   isalice=False\n  if(isalice):\n   alice.append(s)\n  else:\n   bob.append(s)\n ali={}\n bo={}\n for i in alice:\n  d={}\n  for j in i:\n   if(j in d):\n    d[j]+=1\n   else:\n    d[j]=1\n  for j in d:\n   if j not in ali:\n    ali[j]=(1,d[j])\n   else:\n    ali[j]=(ali[j][0]+1,ali[j][1]+d[j])\n for i in bob:\n  d={}\n  for j in i:\n   if(j in d):\n    d[j]+=1\n   else:\n    d[j]=1\n  for j in d:\n   if j not in bo:\n    bo[j]=(1,d[j])\n   else:\n\n    bo[j]=(bo[j][0]+1,bo[j][1]+d[j])\n ans=1\n for i in ali:\n  ans*=ali[i][0]\n for i in bo:\n  ans=ans/bo[i][0]\n x=1\n y=1\n\n for i in bo:\n  x=x*bo[i][1]\n for i in ali:\n  y=y*ali[i][1]\n # print(x,y)\n alice=len(alice)\n bob=len(bob)\n for i in range(bob):\n  while(alice>0 and ans>10000000):\n   ans=ans/y\n   alice-=1\n  ans*=x\n  if(ans>10000000 and alice==0):\n   break\n while(alice>0):\n  ans=ans/y\n  if(ans<1 and alice>100):\n   ans=0\n   break\n  alice-=1\n if(ans>10000000):\n  print("Infinity")\n else:\n  print(ans)\n#AC\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14487