The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum Let $MU$ be the complex bordism spectrum and let $H\mathbb{Z}$ be the Eilenberg-Maclane spectrum.

Is it know what the structure of the complex cobordism cohomology $MU^{*}(H\mathbb{Z})$ is? 

EDIT: What if instead $H\mathbb{Z}$, one consider $H\mathbb{Z}/(p)$ for a prime $p$?
 A: One can prove that $\mathrm{Map}(H\mathbf{F}_p,MU)$ is contractible. We know that $H\mathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = \bigvee_p E_p$, where $E_p = \bigvee_{0\leq n<\infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)
It is, however, not the case that $\mathrm{Map}(H\mathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $H\mathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $\mathrm{Map}(H\mathbf{Z},MU)$ and $\mathrm{Map}(L_E H\mathbf{Z},MU)$. We therefore need to understand $L_E H\mathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} H\mathbf{Z} \simeq H\mathbf{Q}_p$. It therefore suffices to understand $MU^\ast(H\mathbf{Q})$. But $H\mathbf{Q}$ is the colimit of multiplication by $2,3,5,7,\cdots$ on the sphere, so $MU^\ast(H\mathbf{Q})$ admits a description in terms of $\lim^0$ and $\lim^1$ of multiplication by $2,3,5,7,\cdots$ on $\pi_\ast MU$. In particular, the $\lim^1$ term is $\mathrm{Ext}^1_\mathbf{Z}(\mathbf{Q,Z}) \cong \widehat{\mathbf{Z}}/\mathbf{Z}$.
