user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L175
L20:
  mov    0x0(%rbp),%rsi
  lea    D73(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L14:
  lea    -0x1(%rbx),%r12d
  lea    0x0(,%r12,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L156
L212:
  test   %r12d,%r12d
  jle    L30
  lea    -0x2(%rbx),%ecx
  cmp    $0x2,%ecx
  jbe    L150
  mov    %r12d,%edx
  movdqa D35(%rip),%xmm1
  mov    %rbp,%rax
  movdqa D18(%rip),%xmm3
  shr    $0x2,%edx
  movdqa D65(%rip),%xmm2
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rbp,%rdx,1),%rdx
  nopl   (%rax)
L200:
  movdqa %xmm1,%xmm0
  add    $0x10,%rax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L200
  mov    %r12d,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%r12b
  je     L194
L180:
  movslq %eax,%rdx
  lea    0x2(%rax),%esi
  mov    %esi,0x0(%rbp,%rdx,4)
  lea    0x0(,%rdx,4),%rdi
  lea    0x1(%rax),%edx
  cmp    %edx,%r12d
  jle    L194
  lea    0x3(%rax),%edx
  mov    %edx,0x4(%rbp,%rdi,1)
  cmp    %esi,%r12d
  jle    L194
  add    $0x4,%eax
  mov    %eax,0x8(%rbp,%rdi,1)
L194:
  mov    %rbp,%rbx
  lea    0x4(%rbp,%rcx,4),%r13
  mov    %rbp,%rdx
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L189:
  add    $0x4,%rdx
  cmp    %r13,%rdx
  je     L115
L5:
  mov    (%rdx),%eax
  test   %eax,%eax
  je     L189
  lea    -0x2(%rax,%rax,1),%eax
  cmp    %eax,%r12d
  jle    L189
  nopl   0x0(%rax,%rax,1)
L75:
  movslq %eax,%rcx
  movl   $0x0,0x0(%rbp,%rcx,4)
  add    (%rdx),%eax
  cmp    %eax,%r12d
  jg     L75
  add    $0x4,%rdx
  cmp    %r13,%rdx
  jne    L5
L115:
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  lea    D5(%rip),%r12
  call   <printf@plt>
  jmp    L201
  nopw   0x0(%rax,%rax,1)
L182:
  add    $0x4,%rbx
  cmp    %r13,%rbx
  je     L107
L201:
  mov    (%rbx),%esi
  test   %esi,%esi
  je     L182
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r13,%rbx
  jne    L201
L107:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L175:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L14
  jmp    L20
L156:
  lea    D0(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L212
L30:
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L107
L150:
  xor    %eax,%eax
  jmp    L180
```

tool_calls: name: read_data
arguments: data_tag: D73
type_infered: string
name: read_data
arguments: data_tag: D35
type_infered: byte[16]
name: read_data
arguments: data_tag: D18
type_infered: byte[16]
name: read_data
arguments: data_tag: D65
type_infered: byte[16]
name: read_data
arguments: data_tag: D41
type_infered: string
name: read_data
arguments: data_tag: D5
type_infered: string
name: read_data
arguments: data_tag: D0
type_infered: string

tool: D73: Usage: %s NUMBERS


tool: {"D35": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D18": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D65": [0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: D41: 1 

tool: D5: %d 

tool: D0: Could not allocate memory

assistant: ```c
int main(int argc, char *argv[]) {
  int limit, i, j;
  if (argc != 2 || ((limit = atoi(argv[1]))) <= 0) {
    printf("Usage: %s NUMBERS\n", argv[0]);
    exit(1);
  }
  limit = limit - 1;
  int *numbers;
  numbers = (int *)malloc(sizeof(int) * limit);
  if (!numbers) {
    perror("Could not allocate memory");
    exit(1);
  }
  for (i = 0; i < limit; i++) {
    numbers[i] = i + 2;
  }
  for (i = 0; i < limit; i++) {
    if (numbers[i] != 0) {
      for (j = 2 * numbers[i] - 2; j < limit; j += numbers[i])
        numbers[j] = 0;
    }
  }
  printf("1 ");
  for (i = 0; i < limit; i++) {
    if (numbers[i] != 0) {
      printf("%d ", numbers[i]);
    }
  }
  printf("\n");
  free(numbers);
  return 0;
}
```