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

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  lea    D27(%rip),%r14
  push   %r13
  lea    D31(%rip),%r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  jmp    L128
  nopl   0x0(%rax,%rax,1)
L46:
  add    $0x1,%rbx
  add    $0x4,%rbp
  cmp    $0x15,%rbx
  je     L165
L128:
  mov    $0x14,%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    -0x1(%rbx),%rcx
  mov    0x0(%rbp),%esi
  cmp    $0x2,%rcx
  jbe    L97
  movd   %esi,%xmm3
  mov    %rbx,%rax
  pshufd $0x0,%xmm3,%xmm1
  shr    $0x2,%rax
  movdqa %xmm1,%xmm0
  pcmpeqd (%rsp),%xmm0
  pand   D3(%rip),%xmm0
  cmp    $0x1,%rax
  je     L79
  movdqa %xmm1,%xmm2
  pcmpeqd 0x10(%rsp),%xmm2
  psubd  %xmm2,%xmm0
  cmp    $0x2,%rax
  je     L79
  movdqa %xmm1,%xmm2
  pcmpeqd 0x20(%rsp),%xmm2
  psubd  %xmm2,%xmm0
  cmp    $0x3,%rax
  je     L79
  movdqa %xmm1,%xmm2
  pcmpeqd 0x30(%rsp),%xmm2
  psubd  %xmm2,%xmm0
  cmp    $0x5,%rax
  jne    L79
  pcmpeqd 0x40(%rsp),%xmm1
  psubd  %xmm1,%xmm0
L79:
  movdqa %xmm0,%xmm1
  mov    %rbx,%rdx
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffffc,%rdx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%bl
  je     L73
L94:
  cmp    (%rsp,%rdx,4),%esi
  jne    L107
  add    $0x1,%eax
L107:
  lea    0x1(%rdx),%rdi
  cmp    %rcx,%rdi
  ja     L73
  cmp    (%rsp,%rdi,4),%esi
  jne    L153
  add    $0x1,%eax
L153:
  add    $0x2,%rdx
  cmp    %rcx,%rdx
  ja     L73
  cmp    (%rsp,%rdx,4),%esi
  jne    L73
  add    $0x1,%eax
L73:
  cmp    $0x1,%eax
  jne    L46
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x15,%rbx
  jne    L128
L165:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L97:
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L94
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D27
type: string
name: read_data
arguments: label: D31
type: string
name: read_data
arguments: label: D9
type: string
name: read_data
arguments: label: D3
type: byte[16]

tool: D27: E' stato inserito un nuovo numero! %d


tool: D31: Prego insierire il %d numero di %d: 

tool: D9: %d

tool: {"D3": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int arrayNumeriCasuali[20] = {0};
  int temp;
  size_t indice;
  size_t indice2;
  int uguali;
  indice = 0;
  while (indice < 20) {
    printf("Prego insierire il %d numero di %d: ", (indice + 1), 20);
    scanf("%d", &arrayNumeriCasuali[indice]);
    temp = arrayNumeriCasuali[indice];
    uguali = 0;
    indice2 = 0;
    while (indice2 <= indice) {
      if (temp == arrayNumeriCasuali[indice2]) {
        uguali += 1;
      }
      indice2++;
    }
    if (uguali == 1) {
      printf("E' stato inserito un nuovo numero! %d\n", temp);
    }
    indice++;
  }
  return 0;
}
```