Colouring a necklace

The Polya Enumeration Theorem tells us that the number of ways $N$ of colouring the beads can be found by obtaining the cycle index of the dihedral group $D_{11}$ ; note that $Z\left(D_n\right)=\dfrac 12Z\left(C_n\right)+\left\{ \begin{array}{lll} \dfrac 12a_1a_2^{\frac{n-1}{2}} & \hspace{0.5cm} & \text{if }\; n\equiv 1\;\; (\text{mod }2) \\ \dfrac 14\left(a_2^{\frac{n}{2}}+a_1^2a_2^{\frac{n-2}{2}}\right) & & \text{if }\; n\equiv 0\;\; (\text{mod }2), \end{array}\right.$ where $C_n$ is the cyclic group, and $Z\left(C_n\right)=\dfrac 1n \sum_{k\mid n}\phi (k)a_k^{\frac nk}.$ Evaluating this sum at $n=11$ and substituting $a_i=r^i+g^i$ (to account for the two colours) yields the generating function $Z\left(D_{11}\right)_{[r+g]}=r^{11}+g^{11}+r^{10}g+rg^{10}+5r^9g^2+5r^2g^9+10r^8g^3+10r^3g^8+20r^7g^4+20r^4g^7+26r^6g^5+26r^5g^6,$ and thus $N=1+1+1+1+5+5+10+10+20+20+26+26=\boxed{126}.$