An inequality of eigenvalues of $A$ and $U^TAU$, where $A$ is symmetric, $U^TU = I$.

Question

The question is as follows:

$A$ is a $J\times J$ symmetric matrix, $U$ is a $J\times K$ matrix where $K\le J$. Suppose that $U^TU=I_K$. Please show that $\lambda_{j}(U^TAU)\le\lambda_{j}(A)$

I have tried to do eigenvalue decomposition for $A$ and then $U^TAU$ can be described as $P\Lambda P^T$ where $PP^T=I$. However I can't find any fine property. Any hint or solution will be appreciated!