The Moore $($or synchronous$)$ sequence detector circuit has the inputs X$,$ Rx$,$ CLK and the output Z as shown in Figure $1.$ Figure $1$: Port description of sequence detector When the Rx signal goes from $0$ to $1,$ the sequence detector is operational, and the input X is ready to receive a binary signal. When Rx $=0,$ the circuit resets and the output Z is $0.$ The output Z goes to $1$ when X receives the input sequence of $$X$3$X$2$X$1$X$0,$ where $$X$0$ is the first bit of the sequence, followed by$$ X$1$ at the next clock cycle and so on$.$ The output Z will hold its value of $1$ until X receives a second sequence of $$Y$3$Y$2$Y$1$Y$0,$ to make the output Z go to $0.$ Once again, $$Y$0$ is the first bit of the sequence, followed by$$ Y$1$ at the next clock cycle and so on$.$ For all the other inputs of X the output Z will hold its value. The sequences $$X$3$X$2$X$1$X$0$ and $$Y$3$Y$2$Y$1$Y$0$ are determined by using the last three digits of your student number to form a decimal number ranging from $0$ to $999.$ Converting this decimal number to binary yields a maximum of $10$ bits $($Z$1$Z$0$Y$3$Y$2$Y$1$Y$0$X$3$X$2$X$1$X$0)$ where $$X$0$ is the least significant bit and $$Z$1$ is the most significant bit. Let the first four least significant bits be used for $$X$3$X$2$X$1$X$0$ and let the next four bits be used for $$Y$3$Y$2$Y$1$Y$0.$ Bits $$Z$1$Z$0$ can be ignored. If any of the sequences for $$X$3$X$2$X$1$X$0$ and $$Y$3$Y$2$Y$1$Y$0$ are $0000$ then change the sequence to $0001$ and if any of the sequences are $1111$ then then change the sequence to $0111.$

show me the circuit diagram using flip flop, for the decimal number $(070)$