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 = 'import math\nfrom math import log10\n\ndef score(df):\n#     a=""\n n=len(df)\n \n #calulation of fxc\n#     for val in df:\n#         a=a+val     \n  \n \n res = {} \n for val in df:\n  for key in val:\n   if key not in res:\n    res[key]=1\n   else:\n    res[key]+=1\n     \n \n#     print(res)\n fx=1\n for val in res.values():\n  fx=fx*val\n  \n li2={}\n #calculation of xc\n for val in res:\n  li2[val]=0\n  for i in df:\n   if val in i:\n    li2[val]+=1\n \n x=1\n for val in li2:\n  \n  x=x*li2[val]\n#     print(x,fx,"x","fc")\n return [x,fx]\n\n\ndef prog(li):\n al=[]\n bo=[]\n for cur in li:\n  prev = -1\n  f = 0\n  for j in range(len(cur)):\n   if cur[j] not in [\'a\',\'e\',\'i\',\'o\',\'u\']:\n    if prev==-1:\n     prev = j\n    else:\n     if abs(prev-j)==2 or abs(prev-j)==1:\n      f = 1\n      break\n     else:\n      prev = j\n  if f==1:\n   bo.append(cur)\n  else:\n   al.append(cur)\n   \n#     print(al)\n#     print(bo)\n sca=score(al)\n scb=score(bo)\n#     print(sca,scb,"dd")\n ans1 = log10(sca[0])+len(bo)*log10(scb[1])\n ans2 = log10(scb[0])+len(al)*(log10(sca[1]))\n ans1 = ans1-ans2\n#     print(ans1)\n \n return ans1\n \n  \n    \n    \nt = int(input())\nfor i in range(0,t):\n li=[]\n x= int(input())\n for i in range(0,x):\n  e=input()\n  li.append(e)\n final =prog(li)\n if final >7.0:\n  print("Infinity")\n else:\n  print(pow(10,final))'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13369