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

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L108
  lea    0x8(%rsp),%r12
  lea    D27(%rip),%r14
  lea    0x10(%rsp),%rbx
  lea    D9(%rip),%r13
  nopl   (%rax)
L1:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  je     L218
  mov    %rbx,%rdi
  xor    %esi,%esi
  nopl   0x0(%rax)
L165:
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x7,%edx
  sub    %ecx,%edx
  test   %eax,%eax
  mov    %edx,(%rdi)
  lea    0x7(%rax),%edx
  cmovs  %edx,%eax
  mov    %esi,%edx
  add    $0x4,%rdi
  add    $0x1,%esi
  sar    $0x3,%eax
  jne    L165
  mov    %esi,%r9d
  mov    %esi,%edi
  and    $0x1,%r9d
  sar    %edi
  je     L243
  lea    -0x1(%rdi),%ecx
  cmp    $0x2,%ecx
  jbe    L222
  movslq %edx,%rax
  mov    %edi,%ecx
  lea    -0xc(%rbx,%rax,4),%rax
  shr    $0x2,%ecx
  movdqu (%rax),%xmm2
  pshufd $0x1b,%xmm2,%xmm0
  pcmpeqd 0x10(%rsp),%xmm0
  pand   D14(%rip),%xmm0
  cmp    $0x1,%ecx
  je     L78
  movdqu -0x10(%rax),%xmm3
  pshufd $0x1b,%xmm3,%xmm1
  pcmpeqd 0x20(%rsp),%xmm1
  psubd  %xmm1,%xmm0
  cmp    $0x2,%ecx
  je     L78
  movdqu -0x20(%rax),%xmm4
  pshufd $0x1b,%xmm4,%xmm1
  pcmpeqd 0x30(%rsp),%xmm1
  psubd  %xmm1,%xmm0
  cmp    $0x3,%ecx
  je     L78
  movdqu -0x30(%rax),%xmm5
  pshufd $0x1b,%xmm5,%xmm1
  pcmpeqd 0x40(%rsp),%xmm1
  psubd  %xmm1,%xmm0
  cmp    $0x4,%ecx
  je     L78
  movdqu -0x40(%rax),%xmm6
  pshufd $0x1b,%xmm6,%xmm1
  pcmpeqd 0x50(%rsp),%xmm1
  psubd  %xmm1,%xmm0
  cmp    $0x6,%ecx
  jne    L78
  movdqu -0x50(%rax),%xmm1
  pshufd $0x1b,%xmm1,%xmm1
  pcmpeqd 0x60(%rsp),%xmm1
  psubd  %xmm1,%xmm0
L78:
  movdqa %xmm0,%xmm1
  mov    %edi,%ecx
  mov    %edx,%r8d
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  sub    %ecx,%r8d
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %ecx,%edi
  je     L152
L241:
  movslq %ecx,%r11
  movslq %r8d,%r10
  mov    0x10(%rsp,%r11,4),%r11d
  cmp    %r11d,0x10(%rsp,%r10,4)
  jne    L41
  add    $0x1,%eax
L41:
  lea    0x1(%rcx),%r10d
  lea    -0x1(%r8),%r11d
  cmp    %edi,%r10d
  jge    L152
  movslq %r11d,%r11
  movslq %r10d,%r10
  mov    0x10(%rsp,%r11,4),%r11d
  cmp    %r11d,0x10(%rsp,%r10,4)
  jne    L68
  add    $0x1,%eax
L68:
  add    $0x2,%ecx
  sub    $0x2,%r8d
  cmp    %ecx,%edi
  jle    L152
  movslq %r8d,%r8
  movslq %ecx,%rcx
  mov    0x10(%rsp,%r8,4),%edi
  cmp    %edi,0x10(%rsp,%rcx,4)
  jne    L152
  add    $0x1,%eax
L152:
  add    %eax,%eax
  test   %r9d,%r9d
  je     L27
  cmp    %edx,%eax
  je     L218
L3:
  mov    %r13,%rdi
  call   <printf@plt>
L206:
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L1
L108:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L125
  add    $0xe0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L27:
  cmp    %eax,%esi
  jne    L3
  nopl   0x0(%rax)
L218:
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L206
L222:
  mov    %edx,%r8d
  xor    %ecx,%ecx
  jmp    L241
L243:
  test   %r9d,%r9d
  jne    L218
  jmp    L3
L125:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type_infered: string
name: extract_symbolic_data
arguments: tag: D27
type_infered: string
name: extract_symbolic_data
arguments: tag: D9
type_infered: string
name: extract_symbolic_data
arguments: tag: D14
type_infered: byte[16]

tool: D16: %d

tool: D27: 1


tool: D9: -1


tool: {"D14": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
main() {
  int a, b, i, j = 0, c[50], k = 0, l = 0;
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &a);
    i = a;
    while (i != 0) {
      b = i % 8;
      c[j] = b;
      i = i / 8;
      j++;
    }
    for (i = 0, k = j - 1; i < j / 2; i++, k--) {
      if (c[i] == c[k])
        l++;
    }
    if (j % 2 == 0) {
      if (l * 2 == j)
        printf("1\n");
      else
        printf("-1\n");
    } else {
      if ((l * 2) + 1 == j)
        printf("1\n");
      else
        printf("-1\n");
    }
    j = 0;
    k = 0;
    l = 0;
  }
  return 0;
}
```