How to randomly sample 2 indices of True values for each row in a boolean tensor in PyTorch, in the fastest way?

I'm given a PyTorch boolean tensor of size NxK (K>=2), which is guaranteed to have at least 2 True values in each row. For each row, I want to get the indices of 2 random cells with True values.

For example, let's say I have the following tensor:

tensor([[False,  True, False, False,  True],
        [ True,  True,  True, False,  True],
        [ True,  True, False,  True, False],
        [False,  True, False, False,  True],
        [False,  True, False, False,  True]])

So a possible output would be:

tensor([[1, 4],
        [4, 2],
        [0, 3],
        [4, 1],
        [4, 1]])

Because on the first row, we have True value on column 1 and 4, and on the second row we have True value on column 4 and 2, and so on.

Another possible output would be:

tensor([[4, 1],
        [1, 4],
        [0, 1],
        [1, 4],
        [1, 4]])

Because now a different random True values was selected for each row.

I've currently implemented a pytorch-numpy-pytorch solution for that, using np.random.choice:

available_trues = [t.nonzero(as_tuple=False).flatten() for t in input_tensor]
only_2_trues = [np.random.choice(t.cpu(), size=(2,), replace=False) for t in available_trues]
only_2_trues = torch.from_numpy(np.stack(only_2_trues)).cuda()

But this solution is not vectorized (since it works using lists) and it requires moving the data back and forth between the CPU and the GPU. Since I'm working on large matrices and running this operation many times, this causes a major slowdown.

I wonder if it can be done without list comprehension or without moving the data (or even without both :] )