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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movl   $0x1fca055,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  sar    $0x6,%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x2,-0x24(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  mov    %eax,-0x14(%rbp)
  jmp    L26
L79:
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L39
L9:
  mov    -0x20(%rbp),%eax
  sar    $0x6,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  and    $0x3f,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L33
  mov    -0x20(%rbp),%eax
  sar    $0x6,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  and    $0x3f,%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%ecx
  mov    -0x20(%rbp),%eax
  sar    $0x6,%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  or     %ecx,%edx
  mov    %edx,(%rax)
L33:
  mov    -0x24(%rbp),%eax
  add    %eax,-0x20(%rbp)
L39:
  mov    -0x20(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L9
  nop
L76:
  addl   $0x1,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L26
  mov    -0x24(%rbp),%eax
  sar    $0x6,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x24(%rbp),%eax
  and    $0x3f,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L76
L26:
  mov    -0x18(%rbp),%eax
  sar    %eax
  cmp    %eax,-0x24(%rbp)
  jle    L79
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L96
L86:
  mov    -0x1c(%rbp),%eax
  sar    $0x6,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  and    $0x3f,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L106
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L59
L106:
  subl   $0x1,-0x1c(%rbp)
L96:
  cmpl   $0x0,-0x1c(%rbp)
  jg     L86
  mov    $0x0,%eax
L59:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string

tool: D2: %d


assistant: ```c
int main() {
  int limit = 33333333;
  size_t primes_size = ((limit >> 6) + 1) * sizeof(uint64_t);
  uint64_t *primes = (uint64_t *)malloc(primes_size);
  int64_t p = 2, sqrt_limit = (int64_t)sqrt(limit);
  while (p <= limit >> 1) {
    for (int64_t n = 2 * p; n <= limit; n += p)
      if (!(primes[n >> 6] & 1 << (n & 0x3f)))
        (primes[n >> 6] |= 1 << (n & 0x3f));
    while (++p <= sqrt_limit && (primes[p >> 6] & 1 << (p & 0x3f)))
      ;
  }
  for (int i = limit; i > 0; i--) {
    if (!(primes[i >> 6] & 1 << (i & 0x3f))) {
      printf("%d\n", i);
      return 0;
    }
  }
}
```