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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L91:
  lea    D114(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xfc4(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xfc4(%rbp),%eax
  cmp    $0x3e8,%eax
  jg     L91
  mov    -0xfc4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L91
L10:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xfc0(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xfc0(%rbp),%eax
  test   %eax,%eax
  js     L10
  mov    -0xfc0(%rbp),%edx
  mov    -0xfc4(%rbp),%eax
  cmp    %eax,%edx
  jg     L10
L39:
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xfbc(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xfbc(%rbp),%edx
  mov    -0xfc0(%rbp),%eax
  cmp    %eax,%edx
  jl     L39
  mov    -0xfbc(%rbp),%edx
  mov    -0xfc4(%rbp),%eax
  cmp    %eax,%edx
  jg     L39
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xfb8(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0xfb4(%rbp)
  jmp    L78
L41:
  lea    D95(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xfb0(%rbp),%rdx
  mov    -0xfb4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0xfb4(%rbp)
L78:
  mov    -0xfbc(%rbp),%eax
  cmp    %eax,-0xfb4(%rbp)
  jl     L41
  movl   $0x0,-0xfb4(%rbp)
  jmp    L50
L107:
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0xfb8(%rbp),%eax
  imul   %eax,%edx
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xfb4(%rbp)
L50:
  mov    -0xfbc(%rbp),%eax
  cmp    %eax,-0xfb4(%rbp)
  jl     L107
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D114
datatype: string
name: disassemble_data
arguments: tag: D106
datatype: string
name: disassemble_data
arguments: tag: D11
datatype: string
name: disassemble_data
arguments: tag: D14
datatype: string
name: disassemble_data
arguments: tag: D29
datatype: string
name: disassemble_data
arguments: tag: D95
datatype: string
name: disassemble_data
arguments: tag: D43
datatype: string

tool: D114: Inserisci il valore di N2

tool: D106: %d

tool: D11: Inserisci il valore di N1

tool: D14: Inserisci il valore di K

tool: D29: Inserire lo scalare

tool: D95: Inserire il vettore 

tool: D43: PAUSE

assistant: ```c
main() {
  int N2;
  int N1;
  int K;
  int Num;
  int vett[999];
  int I;
  do {
    printf("Inserisci il valore di N2");
    scanf("%d", &N2);
  } while (N2 > 1000 || N2 % 2 == 0);
  do {
    printf("Inserisci il valore di N1");
    scanf("%d", &N1);
  } while (N1 < 0 || N1 > N2);
  do {
    printf("Inserisci il valore di K"), scanf("%d", &K);
  } while (K < N1 || K > N2);
  printf("Inserire lo scalare");
  scanf("%d", &Num);
  for (I = 0; I < K; I++) {
    printf("Inserire il vettore ");
    scanf("%d", &vett[I]);
  }
  for (I = 0; I < K; I++) {
    vett[I] = vett[I] * Num;
    printf("%d", vett[I]);
  }
  system("PAUSE");
}
```