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

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    D8(%rip),%r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x1c8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    -0x1d4(%rbp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  movslq -0x1d4(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L153
L133:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L133
L153:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L161
L90:
  lea    0x3(%rsp),%r12
  mov    %r12,%rax
  and    $0xfffffffffffffffc,%r12
  shr    $0x2,%rax
  mov    %rax,-0x1e8(%rbp)
  test   %ecx,%ecx
  jle    L105
  lea    -0x1d0(%rbp),%r13
  xor    %r15d,%r15d
  mov    %r13,%rbx
  cs nopw 0x0(%rax,%rax,1)
L54:
  mov    %rbx,%rsi
  mov    %r14,%rdi
  add    $0x1,%r15d
  add    $0x4,%rbx
  call   <scanf@plt>
  mov    -0x1d4(%rbp),%r8d
  cmp    %r15d,%r8d
  jg     L54
  test   %r8d,%r8d
  jle    L105
  mov    %r8d,%r9d
  mov    %r12,%rax
  lea    (%r12,%r9,4),%rdx
  nopw   0x0(%rax,%rax,1)
L11:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L11
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L107:
  test   %rcx,%rcx
  je     L39
  mov    0x0(%r13,%rcx,4),%edi
  lea    -0x1(%rcx),%esi
  xor    %eax,%eax
  jmp    L97
  nopl   0x0(%rax)
L155:
  mov    %rdx,%rax
L97:
  cmp    0x0(%r13,%rax,4),%edi
  jle    L84
  mov    (%r12,%rax,4),%edx
  cmp    (%r12,%rcx,4),%edx
  jl     L84
  add    $0x1,%edx
  mov    %edx,(%r12,%rcx,4)
L84:
  lea    0x1(%rax),%rdx
  cmp    %rax,%rsi
  jne    L155
L39:
  add    $0x1,%rcx
  cmp    %r9,%rcx
  jne    L107
  mov    -0x1e8(%rbp),%rax
  mov    0x0(,%rax,4),%esi
  cmp    $0x1,%r8d
  jle    L94
  lea    -0x2(%r8),%edx
  lea    0x4(%r12),%rax
  lea    0x8(%r12,%rdx,4),%rcx
  nopw   0x0(%rax,%rax,1)
L121:
  mov    (%rax),%edx
  cmp    %edx,%esi
  cmovl  %edx,%esi
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L121
L94:
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L168
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L161:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L90
L105:
  mov    -0x1e8(%rbp),%rax
  mov    0x0(,%rax,4),%esi
  jmp    L94
L168:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D8
type_infered: string
name: parse_data
arguments: tag: D16
type_infered: string

tool: D7: Enter num

tool: D8: %d

tool: D16: %d


assistant: ```c
int main() {
  int i, arr[100], num, j;
  printf("Enter num\n");
  scanf("%d", &num);
  int lis[num];
  for (i = 0; i < num; i++)
    scanf("%d", &arr[i]);
  for (i = 0; i < num; ++i)
    lis[i] = 1;
  for (i = 0; i < num; ++i) {
    for (j = 0; j < i; ++j) {
      if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
        lis[i] = lis[j] + 1;
    }
  }
  int max = lis[0];
  for (i = 1; i < num; ++i) {
    if (max < lis[i])
      max = lis[i];
  }
  printf("%d\n", max);
  return 0;
}
```