Some integral numbers are odd. All are more odd, or less odd, than others.
Even numbers satisfy `n = 2m` ( with `m` also integral ) and we will ( completely arbitrarily ) think of odd numbers as `n = 2m + 1`.  
Now, some odd numbers can be more odd than others: when for some `n`, `m` is more odd than for another's. Recursively. :]  
Even numbers are always less odd than odd numbers, but they also can be more, or less, odd than other even numbers, by the same mechanism.
# Task
Given a _non-empty_ finite list of _unique_ integral ( not necessarily non-negative ) numbers, determine the number that is _odder than the rest_.  
Given the constraints, there will always be exactly one such number.
# Examples
```python
oddest([1,2]) => 1
oddest([1,3]) => 3
oddest([1,5]) => 5
```
# Hint
Do you _really_ want one? Point or tap here.
from functools import cmp_to_key

def comp(x, y):
    rx, ry = x&1, y&1
    while rx == ry:
        x, y = (x - rx) >> 1, (y - ry) >> 1
        rx, ry = x&1, y&1
    return rx > ry

def oddest(a):
    return max(a, key=cmp_to_key(comp))
from functools import cmp_to_key

def comp(x, y):
    rx, ry = x&1, y&1
    while rx == ry:
        x, y = (x - rx) >> 1, (y - ry) >> 1
        rx, ry = x&1, y&1
    return rx > ry

def oddest(a):
    return max(a, key=cmp_to_key(comp))