Please enter all answers as strings of 0's and 1's. Consider the 64 bit DES block below. 00100000001000010001000000000000000000000000000000000000000000000010000000100001000100000000000000000000000000000000000000000000 Suppose that this is the initial input to DES. What will the block look like after the initial permutation (IP) is applied? If we apply the IP−1−1 transformation to the above, what will the result be?

This code is correct in reversing a 32-bit integer when looking at the binary form of the integer. But it uses 54 operations instead of 45. We are strictly limited to using only !, +, ^, &, <<, >>, |, ~. We cannot use constants that are anything other than 0x0 to 0xFF. We also cannot use loops or conditionals, in other words "straight line" code. I am unsure where I can simplify this code in order to get it into the required maximum number of operations (45).   Will leave good review!   int bitReverse(int x) {int result, mask1, mask2, mask4, mask8, mask16, min; min = 1 << 31;mask16 = ((0xff << 8) | 0xff);mask8 = (mask16 << 8) ^ mask16;mask4 = (mask8 << 4) ^ mask8;mask2 = (mask4 << 2) ^ mask4;mask1 = (mask2 << 1) ^ mask2;mask1 = ~mask1;mask2 = ~mask2;mask4 = ~mask4;mask8 = ~mask8;mask16 = ~mask16; result = ((x << 16) & mask16) | (((x & mask16) >> 16) & (~mask16));result = ((result << 8) & mask8) |…