user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D44(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D63(%rip),%rbx
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopw   0x0(%rax,%rax,1)
L73:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    (%rsp),%eax
  cmp    $0x3e8,%eax
  jg     L73
  test   $0x1,%al
  je     L73
  lea    D36(%rip),%r12
  lea    0x4(%rsp),%rbp
  nopl   0x0(%rax)
L7:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L7
  cmp    (%rsp),%eax
  jg     L7
  lea    D23(%rip),%r12
  lea    0x8(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  cmp    0x4(%rsp),%eax
  jl     L22
  cmp    (%rsp),%eax
  jg     L22
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L20
  lea    0x10(%rsp),%rbp
  xor    %r12d,%r12d
  lea    D21(%rip),%r14
  mov    %rbp,%r13
  nopl   (%rax)
L32:
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  add    $0x4,%r13
  cmp    %r12d,%eax
  jg     L32
  test   %eax,%eax
  jle    L20
  xor    %r12d,%r12d
  nop
L10:
  mov    0x0(%rbp),%esi
  imul   0xc(%rsp),%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  add    $0x4,%rbp
  mov    %esi,-0x4(%rbp)
  call   <printf@plt>
  cmp    %r12d,0x8(%rsp)
  jg     L10
L20:
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <system@plt>
  mov    0xfb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0xfc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D44
type_guessed: string
name: disassemble_data
arguments: tag: D63
type_guessed: string
name: disassemble_data
arguments: tag: D36
type_guessed: string
name: disassemble_data
arguments: tag: D23
type_guessed: string
name: disassemble_data
arguments: tag: D10
type_guessed: string
name: disassemble_data
arguments: tag: D21
type_guessed: string
name: disassemble_data
arguments: tag: D26
type_guessed: string

tool: D44: Inserisci il valore di N2

tool: D63: %d

tool: D36: Inserisci il valore di N1

tool: D23: Inserisci il valore di K

tool: D10: Inserire lo scalare

tool: D21: Inserire il vettore 

tool: D26: 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");
}
```