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 = 'from math import log10\nimport math\nfor _ in range(int(input())):\n n = int(input())\n alice = []\n bob = []\n words = []\n for i in range(n):\n  words.append(input())\n for cur in words:\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   bob.append(cur)\n  else:\n   alice.append(cur)\n  \n#     print(alice)\n#     print(bob)\n # Alice \n n1 = len(alice)\n freq_a = {}\n for i in alice:\n  for j in i:\n   if j not in freq_a:\n    freq_a[j] = 1\n   else:\n    freq_a[j]+=1\n num_a = {}\n for i in freq_a:\n  num_a[i] = 0\n  for j in alice:\n   if i in j:\n    num_a[i]+=1\n#     print(freq_a)\n#     print(num_a)\n num1 = 1\n den1 = 1\n for i in num_a:\n  num1 = (num1*num_a[i])\n  den1 = den1*freq_a[i]\n \n \n # Bob alphabets\n n2 = len(bob)\n freq_b = {}\n for i in bob:\n  for j in i:\n   if j not in freq_b:\n    freq_b[j] = 1\n   else:\n    freq_b[j]+=1\n num_b = {}\n for i in freq_b:\n  num_b[i] = 0\n  for j in bob:\n   if i in j:\n    num_b[i]+=1\n num2 = 1\n den2 = 1\n for i in num_b:\n  num2 = (num2*num_b[i])\n  den2 = den2*freq_b[i]\n if n<=10:\n#         if n2>=n1:\n#             ans = pow(den2/den1,n1)*pow(den2,n2-n1)\n#         else:\n#             ans = pow(den2/den1,n2)*pow(den1,-(n1-n2))\n#         t = num1/num2\n#         ans = ans*t\n#         if ans>10000000.0:\n#             print("Infinity")\n#         else:\n#             print(ans)\n#     else:\n  ans1 = log10(num1)+n2*log10(den2)\n  ans2 = log10(num2)+n1*(log10(den1))\n  ans1 = ans1-ans2\n  if ans1 > 7.0:\n   print("Infinity")\n  else:\n   print(pow(10,ans1))'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13911