Please solve and show steps, work, and information related to this problem. Thank you. The following C++ code performs a set of various arithmetic operations using both pointers and references:Fill in a memory table that shows the final values for both function a and function b.Indicate the final values of each parameter, passed in variable(s), and the return value of the function. Provide a separate copy of this memory table for each function. #include using std::cout;int function b int pl, int& p2, int* p3);int function a(int pl, int& p2, int p3);int main(){}int{int x, y, Zi.x = 2;y = 3;z = 4;cout << "Call 1: " << function a(&x, y, z) << "\n";cout << "Call 2: " << function_b(y, x, &z) << "\n";function a(int* pl, int& p2, int p3)int apli*p1=p2 p3;p2 = a + p3.;.p3 = 5;int b= *p1 + p2 + p3;return bbint function_b(int pl, int& p2, int* p3)for (int i = p1; i < p2 * 10; i++){*p3 = *p3 + 5:}return pl + p2 + *p3;Fill in a memory table that shows the final values for both function a and function b.Indicate the final values of each parameter, passed in variable(s), and the return value of the function. Provide a separate copy of this memory table for each function.Variable NameAddress0x01Ox020x030x040x050x060x07Return Value0x080x090x10➖➖➖➖➖➖Value

Function a(int *p1, int &p2, int p3) Here p1 is a pointer variable refer to variable 'x' of main(). p2 is a reference variable which means p2 value is same as of variable 'y' of main(). p3 is simple variable that take value of variable 'z' of main(). int a = *p1 ; // *p1 refers x which is 2 assingned to 'a'. *p1 = p2 * p3; //*p1 refers x whose value now become 12. p2 = a + p3; // a+p3 = 6 which is assigned to p2. p3 = 5; int b = *p1 + p2 + p3 ; Memory Table for function a Variable Name Value *p1 12 p2 6 p3 5 a 2 b 23 Return value 23Function b(int p1, int &p2, int *p3) Here p1 is a simple variable refer to variable 'y' of main(). p2 is a reference variable which means p2 value is same as of variable 'x' of main(). p3 is pointer variable that refers to variable 'z' of main(). for(int i = p1 ; i < p2 * 10 ; i++){ *p3 = *p3 + 5; } Here loop will run from i= 6 to 119 because p2 * 10 = 120 and at each iteration p3 that refers variable 'z' of main() will get incremented by 5 so p3 will become 574 finally. Memory Table for function b Variable Name Value p1 6 p2 12 *p3 574 Return value 592