Two tortoises named ***A*** and ***B*** must run a race. ***A*** starts with an average speed of ```720 feet per hour```.
Young ***B*** knows she runs faster than ***A***, and furthermore has not finished her cabbage.
When she starts, at last, she can see that ***A*** has a `70 feet lead` but ***B***'s speed is `850 feet per hour`.
How long will it take ***B*** to catch ***A***?
More generally:
given two speeds `v1` (***A***'s speed, integer > 0) and `v2` (***B***'s speed, integer > 0) and a lead `g` (integer > 0)
how long will it take ***B*** to catch ***A***? 
The result will be an array ```[hour, min, sec]``` which is the time needed in hours, minutes and seconds (round down to the nearest second)
or a string in some languages.
If `v1 >= v2` then return `nil`, `nothing`, `null`, `None` or `{-1, -1, -1}` for C++, C, Go, Nim, `[]` for Kotlin or "-1 -1 -1".
## Examples:
(form of the result depends on the language)
```
race(720, 850, 70) => [0, 32, 18] or "0 32 18"
race(80, 91, 37)   => [3, 21, 49] or "3 21 49"
```
** Note: 
- See other examples in "Your test cases".
- In Fortran - as in any other language - the returned string is not permitted to contain any redundant trailing whitespace: you can use dynamically allocated character strings.
** Hints for people who don't know how to convert to hours, minutes, seconds:
- Tortoises don't care about fractions of seconds
- Think of calculation by hand using only integers (in your code use or simulate integer division)
- or Google: "convert decimal time to hours minutes seconds"
def race(v1, v2, g):
    if v1 >= v2: return None
    t = float(g)/(v2-v1)*3600
    mn, s = divmod(t, 60)
    h, mn = divmod(mn, 60)
    return [int(h), int(mn), int(s)]
def race(v1, v2, g):
    if v1 >= v2: return None
    t = float(g)/(v2-v1)*3600
    mn, s = divmod(t, 60)
    h, mn = divmod(mn, 60)
    return [int(h), int(mn), int(s)]