Calculate power for int64 exponent precisely

[hcirellu]

I added a precise calculation for POW64 with an overflow detection.

The example from #288 now returns the correct values.

x = as.integer64(c("94906267", "2147483650"))
> x^2L
integer64
[1] 9007199515875289    4611686027017322500

The results for double exponent are unchanged, i.e.:

x = as.integer64(c("94906267", "2147483650"))
> x^2
integer64
[1] 9007199515875288    4611686027017322496

In addition, the new implementation is faster, too.

x = as.integer64(sample(-100:100,rep=T,1000000))
y32 = 0:9
yd = as.double(y32)
microbenchmark::microbenchmark(x^y32, x^yd)
# Unit: milliseconds
#   expr       min        lq     mean    median        uq      max neval
#  x^y32  6.262101  6.566502 13.84232  6.816252  7.256201 178.3225   100
#   x^yd 27.294501 27.965501 30.46499 28.709051 31.592151  66.1929   100

Closes #288