user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x1,%esi
  lea    D10(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  movdqa D21(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D15(%rip),%rax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  movdqa D20(%rip),%xmm0
  mov    %rbp,%rbx
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L250:
  mov    %r12d,%esi
L25:
  mov    0x8(%rbx),%ecx
  mov    0x4(%rbx),%edx
  lea    0x1(%rsi),%r12d
  mov    %r12d,%eax
  cmp    %ecx,%edx
  jle    L64
  lea    -0x1(%rsi),%eax
  mov    %edx,0x8(%rbx)
  test   %eax,%eax
  je     L12
  mov    (%rbx),%edx
  cmp    %edx,%ecx
  jge    L76
  mov    %esi,%edi
  mov    %edx,0x4(%rbx)
  sub    $0x2,%edi
  je     L12
  mov    -0x4(%rbx),%edx
  cmp    %edx,%ecx
  jge    L64
  mov    %esi,%eax
  mov    %edx,(%rbx)
  sub    $0x3,%eax
  je     L12
  mov    -0x8(%rbx),%edx
  cmp    %edx,%ecx
  jge    L4
  mov    %esi,%edi
  mov    %edx,-0x4(%rbx)
  sub    $0x4,%edi
  je     L12
  mov    -0xc(%rbx),%edx
  cmp    %edx,%ecx
  jge    L64
  mov    %esi,%r8d
  mov    %edx,-0x8(%rbx)
  sub    $0x5,%r8d
  je     L12
  mov    -0x10(%rbx),%eax
  cmp    %eax,%ecx
  jge    L4
  mov    %eax,-0xc(%rbx)
  mov    %esi,%eax
  sub    $0x6,%eax
  je     L12
  mov    -0x14(%rbx),%edx
  cmp    %edx,%ecx
  jge    L31
  mov    %edx,-0x10(%rbx)
  cmp    $0x7,%esi
  je     L12
  mov    -0x18(%rbx),%edx
  cmp    %edx,%ecx
  jge    L64
  mov    %edx,-0x14(%rbx)
  cmp    $0x8,%esi
  je     L194
  mov    -0x1c(%rbx),%eax
  cmp    %ecx,%eax
  jle    L29
  mov    %eax,-0x18(%rbx)
  mov    $0x1,%eax
L64:
  cltq
  mov    %r13,%rdi
  add    $0x4,%rbx
  mov    %ecx,(%rsp,%rax,4)
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <imprimir_arreglo@plt>
  cmp    $0x9,%r12d
  jne    L250
L254:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  mov    %r13,%rdi
  mov    %ecx,0x4(%rsp)
  add    $0x4,%rbx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <imprimir_arreglo@plt>
  jmp    L250
L4:
  mov    %edi,%eax
  jmp    L64
L76:
  mov    %esi,%eax
  jmp    L64
L31:
  mov    %r8d,%eax
  jmp    L64
L194:
  mov    %r13,%rdi
  mov    $0x8,%esi
  mov    %ecx,0x4(%rsp)
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <imprimir_arreglo@plt>
  jmp    L254
L29:
  mov    $0x1,%eax
  jmp    L64
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
type: string
name: parse_data
arguments: tag: D21
type: byte[16]
name: parse_data
arguments: tag: D15
type: byte[8]
name: parse_data
arguments: tag: D20
type: byte[16]

tool: D10: Iteracion #%03d 

tool: {"D21": [0x01,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x14,0x00,0x00,0x00,0x0a,0x00,0x00,0x00]}

tool: {"D15": [0x15,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D20": [0x03,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x0e,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int arreglo[10] = {1, 4, 20, 10, 3, 8, 9, 14, 21, 6};
  int i;
  int j;
  int temporal;
  int iteracion = 1;
  for (i = 2; i < 10; i++) {
    temporal = arreglo[i];
    j = i - 1;
    while ((j >= 1) && (arreglo[j] > temporal)) {
      arreglo[j + 1] = arreglo[j];
      j--;
    }
    arreglo[j + 1] = temporal;
    printf("Iteracion #%03d ", iteracion++);
    imprimir_arreglo(arreglo);
  }
  return (0);
}
```