# How do you turn the CX gate upside down on ibm-q-experience?

I am learning how to make Grover's algorithm but the last cx gate needs to be the opposite way around. It always makes the top qubit the control, so how do you turn it around?

$$(H \otimes H) CNOT (H \otimes H),$$
where $$H$$ is Hadamard gate and $$CNOT$$ is controlled NOT with control qubit upside and target qubit downside.