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 time\nvowels = \'aeiou\'\nstart_time = time.time()\ndef main():\n import math as m\n ryt = m.log(10**7, 2)\n lyt = m.log(0.0000009, 2)\n t = int(input())\n v = {"a": 1, "e": 1, "i": 1, "o": 1, "u": 1}\n for qwer in range(t):\n  l = int(input())\n  a = []\n  b = []\n  for wer in range(l):\n   s = input()\n   n = len(s)\n   c1 = 0\n   for i in range(n):\n    if s[i] not in v:\n     if i+1 < n:\n      if s[i+1] not in v:\n       c1 = 1\n       break\n     if i+2 < n:\n      if s[i+2] not in v:\n       c1 = 1\n       break\n   if c1 == 1:\n    b.append(s)\n   else:\n    a.append(s)\n  # for _ in range(l):\n  #     flag = False\n  #     s = input()\n  #     for i in range(len(s)-2):\n  #         if s[i] not in vowels:\n  #             if s[i+1] not in vowels or s[i+2] not in vowels:\n  #                 flag = True\n  #                 break\n  #         else:\n  #             if s[i+1] not in vowels and s[i+2] not in vowels:\n  #                 flag = True \n  #                 break\n  #     if flag:\n  #         b.append(s)\n  #     else:\n  #         a.append(s)\n  xa = {}\n  fxa = {}\n  xb = {}\n  fxb = {}\n  ka = len(a)\n  kb = len(b)\n  for i in range(ka):\n   d = {}\n   for j in range(len(a[i])):\n    if a[i][j] in d:\n     d[a[i][j]] += 1\n    else:\n     d[a[i][j]] = 1\n   for u in d:\n    if u in xa:\n     xa[u] += 1\n     fxa[u] += d[u]\n    else:\n     xa[u] = 1\n     fxa[u] = d[u]\n  for i1 in range(kb):\n   d = {}\n   for j1 in range(len(b[i1])):\n    if b[i1][j1] in d:\n     d[b[i1][j1]] += 1\n    else:\n     d[b[i1][j1]] = 1\n   for u1 in d:\n    if u1 in xb:\n     xb[u1] += 1\n     fxb[u1] += d[u1]\n    else:\n     xb[u1] = 1\n     fxb[u1] = d[u1]\n  ans = 0\n  for y in xa:\n   ans += m.log(xa[y], 2)-ka*(m.log(fxa[y], 2))\n  for r in xb:\n   ans += kb*(m.log(fxb[r], 2))-m.log(xb[r], 2)\n  if ans > ryt:\n   print("Infinity")\n  elif ans < lyt:\n   print(0)\n  else:\n   print(2**ans)\n\n\nmain()\n# print(time.time()-start_time)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 15313