user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D16(%rip),%rdi
  push   %r14
  lea    D53(%rip),%r14
  push   %r13
  lea    D34(%rip),%r13
  push   %r12
  lea    D26(%rip),%r12
  push   %rbp
  lea    D38(%rip),%rbp
  push   %rbx
  lea    D35(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D36(%rip),%rdi
  call   <scanf@plt>
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L9:
  pxor   %xmm1,%xmm1
  mov    %r15d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  cvtss2sd (%rsp),%xmm1
  cvtss2sd %xmm0,%xmm0
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm0,(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm0
  movsd  D105(%rip),%xmm3
  movapd %xmm0,%xmm2
  subsd  D50(%rip),%xmm2
  mulsd  %xmm1,%xmm3
  mulsd  %xmm3,%xmm2
  movsd  D50(%rip),%xmm3
  mulsd  %xmm1,%xmm3
  addsd  %xmm3,%xmm2
  cvtsd2ss %xmm2,%xmm2
  cvtss2sd %xmm2,%xmm2
L24:
  mov    %rbx,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x14(%rsp)
  test   %eax,%eax
  jle    L10
L36:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_int@plt>
  mov    %r13,%rdi
  mov    %eax,%r15d
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_float@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  movss  %xmm0,(%rsp)
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_float@plt>
  comiss D75(%rip),%xmm0
  ja     L9
  mov    %r15d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  movss  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movss  (%rsp),%xmm4
  movss  0x8(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  movaps %xmm4,%xmm2
  cvtss2sd %xmm4,%xmm1
  mulss  %xmm0,%xmm2
  cvtss2sd %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm2
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D16
data_type: string
name: retrieve_data
arguments: tag: D53
data_type: string
name: retrieve_data
arguments: tag: D34
data_type: string
name: retrieve_data
arguments: tag: D26
data_type: string
name: retrieve_data
arguments: tag: D38
data_type: string
name: retrieve_data
arguments: tag: D35
data_type: string
name: retrieve_data
arguments: tag: D36
data_type: string
name: retrieve_data
arguments: tag: D105
data_type: f64
name: retrieve_data
arguments: tag: D50
data_type: f64
name: retrieve_data
arguments: tag: D75
data_type: byte[4]

tool: D16: How many people: 

tool: D53: 
Input ID number: 

tool: D34: Input rate of pay: 

tool: D26: Input hours: 

tool: D38: 
ID Number: %d


tool: D35: You've worked %.2f hours, with a rate of $%.2f per hour which equals to a pay of: $%.2f.



tool: D36: %d

tool: D105: 1.5

tool: D50: 40

tool: {"D75": [0x00,0x00,0x20,0x42]}

assistant: ```c
int main(int argc, char **argv) {
  int id_number, num_people;
  float hours, rate, pay;
  printf("How many people: ");
  scanf("%d", &num_people);
  do {
    printf("\nInput ID number: ");
    id_number = get_int();
    printf("Input rate of pay: ");
    rate = get_float();
    printf("Input hours: ");
    hours = get_float();
    if (hours > 40.0) {
      pay = (40.0 * rate) + ((hours - 40.0) * (rate * 1.5));
    } else {
      pay = hours * rate;
    }
    printf("\nID Number: %d\n", id_number);
    printf("You've worked %.2f hours, with a rate of $%.2f per hour "
           "which equals to a pay of: $%.2f.\n\n",
           hours, rate, pay);
    num_people--;
  } while (num_people > 0);
}
```