user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D30(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L136
  lea    -0x1(%rsi),%eax
  cmp    $0xb,%eax
  jbe    L162
  mov    %esi,%edx
  pxor   %xmm4,%xmm4
  movdqa D38(%rip),%xmm5
  xor    %eax,%eax
  movdqa D8(%rip),%xmm6
  shr    $0x2,%edx
  movdqa %xmm4,%xmm3
  xchg   %ax,%ax
L150:
  movdqa %xmm5,%xmm2
  add    $0x1,%eax
  paddd  %xmm6,%xmm5
  movdqa %xmm2,%xmm0
  movdqa %xmm2,%xmm1
  paddd  %xmm2,%xmm4
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm0,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  paddd  %xmm1,%xmm3
  cmp    %eax,%edx
  jne    L150
  movdqa %xmm4,%xmm0
  mov    %esi,%edx
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%edx
  paddd  %xmm0,%xmm4
  lea    0x1(%rdx),%eax
  movdqa %xmm4,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm4
  movdqa %xmm3,%xmm0
  psrldq $0x8,%xmm0
  movd   %xmm4,%ebx
  paddd  %xmm0,%xmm3
  movdqa %xmm3,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm3
  movd   %xmm3,%ebp
  cmp    %edx,%esi
  je     L101
L143:
  mov    %eax,%edx
  add    %eax,%ebx
  imul   %eax,%edx
  add    %edx,%ebp
  lea    0x1(%rax),%edx
  cmp    %esi,%edx
  jg     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x2(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x3(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x4(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x5(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x6(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x7(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x8(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0x9(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  lea    0xa(%rax),%edx
  add    %ecx,%ebp
  cmp    %edx,%esi
  jl     L101
  mov    %edx,%ecx
  add    $0xb,%eax
  add    %edx,%ebx
  imul   %edx,%ecx
  add    %ecx,%ebp
  cmp    %eax,%esi
  jl     L101
  mov    %eax,%edx
  add    %eax,%ebx
  imul   %eax,%edx
  add    %edx,%ebp
  nopl   0x0(%rax,%rax,1)
L101:
  imul   %ebx,%ebx
  lea    D36(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D14(%rip),%rdi
  sub    %ebx,%ebp
  mov    %ebp,%esi
  neg    %esi
  cmovs  %ebp,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L71
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L136:
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  jmp    L101
L162:
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  mov    $0x1,%eax
  jmp    L143
L71:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D30
type_infered: string
name: disassemble_data
arguments: label: D28
type_infered: string
name: disassemble_data
arguments: label: D38
type_infered: byte[16]
name: disassemble_data
arguments: label: D8
type_infered: byte[16]
name: disassemble_data
arguments: label: D36
type_infered: string
name: disassemble_data
arguments: label: D14
type_infered: string

tool: D30: Please insert the upper bound:

tool: D28: %d

tool: {"D38": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D8": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D36: The difference between the sum of the squares of the first %d natural numbers and the square of the sum is:


tool: D14: %d


assistant: ```c
int main() {
  int meja;
  printf("Please insert the upper bound:\n");
  scanf("%d", &meja);
  int i;
  int sumOfSquares = 0;
  int squareOfSum = 0;
  for (i = 1; i <= meja; i++) {
    sumOfSquares += i * i;
    squareOfSum += i;
  }
  squareOfSum *= squareOfSum;
  printf("The difference between the sum of the squares of the first %d natural numbers and the square of the sum is:\n", meja);
  printf("%d\n", abs(sumOfSquares - squareOfSum));
  return 0;
}
```