user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1e,-0x88(%rbp)
  movl   $0xf,-0x84(%rbp)
  movl   $0x0,-0x8c(%rbp)
  jmp    L27
L8:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x8c(%rbp),%eax
  cltq
  mov    %edx,-0x80(%rbp,%rax,4)
  addl   $0x1,-0x8c(%rbp)
L27:
  mov    -0x8c(%rbp),%eax
  cmp    -0x88(%rbp),%eax
  jl     L8
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x88(%rbp),%eax
  lea    -0x1(%rax),%edx
  lea    -0x80(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_list@plt>
  mov    -0x88(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x84(%rbp),%edx
  lea    -0x80(%rbp),%rax
  mov    %edx,%ecx
  mov    %esi,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <find_kth_element_r@plt>
  mov    -0x84(%rbp),%eax
  cltq
  mov    -0x80(%rbp,%rax,4),%edx
  mov    -0x84(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x8c(%rbp)
  jmp    L41
L10:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x8c(%rbp),%eax
  cltq
  mov    %edx,-0x80(%rbp,%rax,4)
  addl   $0x1,-0x8c(%rbp)
L41:
  mov    -0x8c(%rbp),%eax
  cmp    -0x88(%rbp),%eax
  jl     L10
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x88(%rbp),%eax
  lea    -0x1(%rax),%edx
  lea    -0x80(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_list@plt>
  mov    -0x88(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x84(%rbp),%edx
  lea    -0x80(%rbp),%rax
  mov    %edx,%ecx
  mov    %esi,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <find_kth_element_nr@plt>
  mov    -0x84(%rbp),%eax
  cltq
  mov    -0x80(%rbp,%rax,4),%edx
  mov    -0x84(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D22
type_guessed: string
name: read_data
arguments: tag: D15
type_guessed: string
name: read_data
arguments: tag: D12
type_guessed: string

tool: D22: first list (solve recursively):

tool: D15: 
element in position %d is: %d


tool: D12: second list (solve non-recursively):

assistant: ```c
int main() {
  int l[30];
  int size = 30;
  int k = 15;
  int c;
  for (c = 0; c < size; c++)
    l[c] = rand() % 100;
  printf("first list (solve recursively):\n");
  print_list(l, size - 1);
  find_kth_element_r(l, 0, size - 1, k);
  printf("\nelement in position %d is: %d\n", k, l[k]);
  for (c = 0; c < size; c++)
    l[c] = rand() % 100;
  printf("second list (solve non-recursively):\n");
  print_list(l, size - 1);
  find_kth_element_nr(l, 0, size - 1, k);
  printf("\nelement in position %d is: %d\n", k, l[k]);
  return 0;
}
```