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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x54(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x54(%rbp),%eax
  test   %eax,%eax
  jns    L86
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  jmp    L84
L86:
  movl   $0x0,-0x4c(%rbp)
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L30:
  cmp    %rdx,%rsp
  je     L69
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L30
L69:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L78
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L78:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x40(%rbp)
  movl   $0x2,-0x50(%rbp)
  jmp    L19
L13:
  mov    -0x50(%rbp),%eax
  mov    %eax,%edi
  call   <isPrime@plt>
  test   %rax,%rax
  je     L31
  mov    -0x4c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4c(%rbp)
  mov    -0x40(%rbp),%rdx
  cltq
  mov    -0x50(%rbp),%ecx
  mov    %ecx,(%rdx,%rax,4)
L31:
  addl   $0x1,-0x50(%rbp)
L19:
  mov    -0x54(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jl     L13
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x50(%rbp)
  jmp    L96
L39:
  mov    -0x40(%rbp),%rax
  mov    -0x50(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x50(%rbp)
L96:
  mov    -0x54(%rbp),%eax
  cmp    %eax,-0x50(%rbp)
  jl     L39
  mov    $0x0,%eax
L84:
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L75
  call   <__stack_chk_fail@plt>
L75:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D42
type_infered: string
name: retrieve_data
arguments: data_tag: D35
type_infered: string
name: retrieve_data
arguments: data_tag: D4
type_infered: string
name: retrieve_data
arguments: data_tag: D5
type_infered: string
name: retrieve_data
arguments: data_tag: D28
type_infered: string

tool: D42: How many prime numbers? 

tool: D35: %i

tool: D4: 

  A minimum of one prime number is required.

tool: D5: 


tool: D28: %i 

assistant: ```c
int main(void) {
  int i, n, count;
  printf("How many prime numbers? ");
  scanf("%i", &n);
  if (n < 0) {
    printf("\n\n  A minimum of one prime number is required.\n");
    return 0;
  }
  count = 0;
  int prime[n];
  for (i = 2; count < n; ++i) {
    if (isPrime(i))
      prime[count++] = i;
  }
  printf("\n\n");
  for (i = 0; i < n; ++i)
    printf("%i ", prime[i]);
  return 0;
}
```