Macdonald polynomial Grassmannian vanishing conjecture for $U(m)\times U(n)$

Let $m$ and $n$ be integers with $0\leq m\leq n$, let $\mu\overline{\nu}$ be a dominant weight of $U(m+n)$, and let $P_{\mu\overline{\nu}}$ be the corresponding Macdonald polynomial. Here $Z$ normalizes the integral so that the integral of the trivial polynomial is $1$. Grassmannian vanishing conjecture.

$$\frac{1}{Z}\int P_{\mu\overline{\nu}}(x_1,\dots,x_m,y_1,\dots,y_n)\prod_{1\leq i\ne j\leq m}\frac{(x_i/x_j;q)}{(tx_i/x_j;q)}\prod_{1\leq i\ne j\leq n}\frac{(y_i/y_j;q)}{(ty_i/y_j;q)}\prod_{i=1}^m\frac{dx_i}{2\pi\sqrt{-1}x_i}\prod_{i=1}^n\frac{dy_i}{2\pi\sqrt{-1}y_i}=0$$

unless $\mu=\nu$ and $\ell(\mu)\leq m$; in that case the integral equals

$$\frac{C^-_\mu(q;q,t)C^+_\mu(t^{m+n-2}q;q,t)C^0_\mu(t^n,t^m;q,t)}{C^-_\mu(t;q,t)C^+_\mu(t^{m+n-2}t;q,t)C^0_\mu(qt^{n-1},qt^{m-1};q,t)}.$$

The claim is motivated by the Schur integral for the symmetric space $U(m+n)/(U(m)\times U(n))$; the nonzero value was guessed from low-order examples and remains conjectural.