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

```asm
<otp_keystring2keyblock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x200,-0x18(%rbp)
  jbe    L126
  mov    $0xffffffff,%eax
  jmp    L101
L126:
  movl   $0x0,-0x20(%rbp)
  jmp    L76
L77:
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L176
L33:
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  cmp    $0x2f,%eax
  jbe    L72
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  cmp    $0x39,%eax
  jbe    L178
L72:
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  cmp    $0x40,%eax
  jbe    L29
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  cmp    $0x46,%eax
  jbe    L178
L29:
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  cmp    $0x60,%eax
  jbe    L55
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  cmp    $0x66,%eax
  jbe    L178
L55:
  mov    $0xffffffff,%eax
  jmp    L101
L178:
  nop
  addl   $0x1,-0x1c(%rbp)
L176:
  cmpl   $0x1,-0x1c(%rbp)
  jle    L33
  mov    -0x10(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    $0x9,%eax
  jbe    L150
  mov    -0x10(%rbp),%eax
  cmp    $0x16,%eax
  jbe    L34
  mov    -0x10(%rbp),%eax
  sub    $0x27,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L150
L34:
  mov    -0x10(%rbp),%eax
  sub    $0x7,%eax
  mov    %eax,-0x10(%rbp)
L150:
  mov    -0xc(%rbp),%eax
  cmp    $0x9,%eax
  jbe    L81
  mov    -0xc(%rbp),%eax
  cmp    $0x16,%eax
  jbe    L127
  mov    -0xc(%rbp),%eax
  sub    $0x27,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L81
L127:
  mov    -0xc(%rbp),%eax
  sub    $0x7,%eax
  mov    %eax,-0xc(%rbp)
L81:
  mov    -0x10(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  shl    $0x4,%ecx
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%edx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    -0x20(%rbp),%ecx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  add    %esi,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x20(%rbp)
L76:
  mov    -0x20(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  shr    %rdx
  cmp    %rdx,%rax
  jb     L77
  mov    -0x18(%rbp),%rax
  shr    %rax
L101:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L54
  call   <__stack_chk_fail@plt>
L54:
  leave
  ret
```

assistant: ```c
ssize_t otp_keystring2keyblock(const char *s, unsigned char keyblock[256]) {
  unsigned i;
  size_t l = strlen(s);
  if (l > 256 * 2)
    return -1;
  for (i = 0; i < l / 2; ++i) {
    unsigned int n[2];
    int j;
    n[0] = *s++;
    n[1] = *s++;
    for (j = 0; j < 2; ++j) {
      if ((n[j] >= '0' && n[j] <= '9') || (n[j] >= 'A' && n[j] <= 'F') || (n[j] >= 'a' && n[j] <= 'f'))
        continue;
      return -1;
    }
    n[0] -= '0';
    n[1] -= '0';
    if (n[0] > 9) {
      if (n[0] > 'F' - '0')
        n[0] -= 'a' - '9' - 1;
      else
        n[0] -= 'A' - '9' - 1;
    }
    if (n[1] > 9) {
      if (n[1] > 'F' - '0')
        n[1] -= 'a' - '9' - 1;
      else
        n[1] -= 'A' - '9' - 1;
    }
    keyblock[i] = n[0] << 4;
    keyblock[i] += n[1];
  }
  return l / 2;
}
```