Example 1: Take $n=7, h=2$. It is easy to see that $2^{3}=8 \equiv 1(\bmod 7)$, and for $1 < k < 2$, we have $2^{k} \not\equiv 1(\bmod 7)$, so the order of 2 modulo 7 is 3. For $h=-2$, it is easy to calculate that $(-2)^{6}=64 \equiv 1(\bmod 7)$, and for $1 < k < 5$, we have $(-2)^{k} \not\equiv 1(\bmod 7)$, so the order of -2 modulo 7 is 6.