user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x42c(%rbp)
  mov    -0x42c(%rbp),%eax
  mov    %eax,-0x420(%rbp)
  jmp    L11
L39:
  mov    -0x42c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x42c(%rbp)
  mov    -0x420(%rbp),%edx
  cltq
  mov    %dl,-0x3f0(%rbp,%rax,1)
L11:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x420(%rbp)
  cmpl   $0xa,-0x420(%rbp)
  jne    L39
  mov    -0x42c(%rbp),%eax
  cltq
  movb   $0x0,-0x3f0(%rbp,%rax,1)
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x41c(%rbp)
  movl   $0x0,-0x428(%rbp)
  movl   $0x10,-0x410(%rbp)
  movl   $0x8,-0x40c(%rbp)
  movl   $0x6,-0x408(%rbp)
  movl   $0x1,-0x404(%rbp)
  movl   $0x3,-0x400(%rbp)
  movl   $0x5,-0x3fc(%rbp)
  movl   $0x6,-0x3f8(%rbp)
  movq   $0x7,-0x418(%rbp)
  movl   $0x0,-0x424(%rbp)
  movl   $0x6,-0x428(%rbp)
  jmp    L79
L25:
  mov    -0x428(%rbp),%eax
  cltq
  mov    -0x410(%rbp,%rax,4),%eax
  mov    %eax,-0x424(%rbp)
  movl   $0x0,-0x42c(%rbp)
  jmp    L3
L14:
  mov    -0x42c(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  movsbl %al,%eax
  lea    -0x20(%rax),%ecx
  mov    -0x424(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x424(%rbp)
  mov    %eax,%esi
  lea    0x0(,%rax,4),%edx
  mov    %esi,%eax
  sub    %edx,%eax
  add    %ecx,%eax
  movslq %eax,%rdx
  imul   $0xffffffffae4c415d,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x5e,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    $0x20,%eax
  mov    %eax,%edx
  mov    -0x42c(%rbp),%eax
  cltq
  mov    %dl,-0x3f0(%rbp,%rax,1)
  mov    -0x42c(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x1f,%al
  jg     L29
  mov    -0x42c(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  add    $0x5e,%eax
  mov    %eax,%edx
  mov    -0x42c(%rbp),%eax
  cltq
  mov    %dl,-0x3f0(%rbp,%rax,1)
L29:
  addl   $0x1,-0x42c(%rbp)
L3:
  mov    -0x42c(%rbp),%eax
  cmp    -0x41c(%rbp),%eax
  jl     L14
  subl   $0x1,-0x428(%rbp)
L79:
  cmpl   $0x0,-0x428(%rbp)
  jns    L25
  mov    -0x42c(%rbp),%eax
  cltq
  movb   $0x0,-0x3f0(%rbp,%rax,1)
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L67
  call   <__stack_chk_fail@plt>
L67:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
datatype: string

tool: D9: %s


assistant: ```c
int main() {
  char input[1000];
  int ch, i;
  ch = i = 0;
  while ((ch = getchar()) != '\n')
    input[i++] = ch;
  input[i] = '\0';
  int len = strlen(input);
  int key_posi = 0, key[] = {16, 8, 6, 1, 3, 5, 6};
  size_t key_size = sizeof(key) / sizeof(int);
  char output[1000];
  int key_current = 0;
  for (key_posi = 6; key_posi >= 0; key_posi--) {
    key_current = key[key_posi];
    for (i = 0; i < len; i++) {
      input[i] = (((input[i] - 32) - (3 * key_current++)) % 94) + 32;
      if (input[i] < 32)
        input[i] = input[i] + 94;
    }
  }
  input[i] = '\0';
  printf("%s\n", input);
  return 0;
}
```