user: 下面的汇编指令对应的C源代码是什么：

```asm
<sumEvenAfterQueries>:
  endbr64
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rcx,4),%edi
  sub    $0x8,%rsp
  mov    %ecx,(%r9)
  movl   $0x2,(%r8)
  call   <malloc@plt>
  mov    %rax,%r9
  test   %r12d,%r12d
  jle    L77
  lea    -0x1(%r12),%eax
  cmp    $0x2,%eax
  jbe    L28
  mov    %r12d,%edx
  pxor   %xmm1,%xmm1
  movdqa D0(%rip),%xmm3
  mov    %rbx,%rax
  shr    $0x2,%edx
  movdqa %xmm1,%xmm2
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rbx,%rdx,1),%rdx
  nop
L40:
  movdqu (%rax),%xmm0
  movdqu (%rax),%xmm4
  add    $0x10,%rax
  pand   %xmm3,%xmm0
  pcmpeqd %xmm2,%xmm0
  pand   %xmm4,%xmm0
  paddd  %xmm0,%xmm1
  cmp    %rdx,%rax
  jne    L40
  movdqa %xmm1,%xmm0
  mov    %r12d,%edx
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%edx
  paddd  %xmm0,%xmm1
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm1
  movd   %xmm1,%eax
  test   $0x3,%r12b
  je     L129
L33:
  movslq %edx,%rcx
  lea    0x0(,%rcx,4),%rsi
  mov    (%rbx,%rcx,4),%ecx
  lea    (%rax,%rcx,1),%edi
  and    $0x1,%ecx
  lea    0x1(%rdx),%ecx
  cmove  %edi,%eax
  cmp    %ecx,%r12d
  jle    L129
  mov    0x4(%rbx,%rsi,1),%ecx
  lea    (%rax,%rcx,1),%edi
  and    $0x1,%ecx
  cmove  %edi,%eax
  add    $0x2,%edx
  cmp    %edx,%r12d
  jle    L129
  mov    0x8(%rbx,%rsi,1),%edx
  lea    (%rax,%rdx,1),%ecx
  and    $0x1,%edx
  cmove  %ecx,%eax
L129:
  movslq %r13d,%r10
  xor    %edx,%edx
  test   %r13d,%r13d
  jg     L26
  jmp    L91
  nopl   0x0(%rax,%rax,1)
L112:
  mov    %eax,%ecx
  add    %eax,%esi
  sub    %edi,%ecx
  test   %r8d,%r8d
  mov    %ecx,%eax
  cmove  %esi,%eax
  mov    %eax,(%r9,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r10
  je     L91
L26:
  mov    0x0(%rbp,%rdx,8),%rcx
  mov    (%rcx),%esi
  movslq 0x4(%rcx),%rcx
  lea    (%rbx,%rcx,4),%r8
  mov    (%r8),%edi
  lea    (%rdi,%rsi,1),%ecx
  mov    %ecx,(%r8)
  mov    %ecx,%r8d
  and    $0x1,%r8d
  test   $0x1,%dil
  je     L112
  add    %eax,%ecx
  test   %r8d,%r8d
  cmove  %ecx,%eax
  mov    %eax,(%r9,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r10
  jne    L26
L91:
  add    $0x8,%rsp
  mov    %r9,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L77:
  xor    %eax,%eax
  jmp    L129
L28:
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L33
```

tool_calls: name: parse_data
arguments: tag: D0
type_guessed: byte[16]

tool: {"D0": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int *sumEvenAfterQueries(int *A, int ASize, int **queries, int queriesSize, int *queriesColSize, int *returnSize) {
  *returnSize = queriesSize;
  *queriesColSize = 2;
  int *answer = malloc(sizeof(int) * queriesSize);
  int sum = 0;
  for (int i = 0; i < ASize; i++) {
    if (A[i] % 2 == 0) {
      sum += A[i];
    }
  }
  int j, k, l;
  for (int i = 0; i < queriesSize; i++) {
    j = queries[i][1];
    k = queries[i][0];
    l = A[j];
    A[j] += k;
    if (l % 2 == 0) {
      if (A[j] % 2 != 0) {
        sum -= l;
      } else {
        sum += k;
      }
    } else {
      if (A[j] % 2 == 0) {
        sum += A[j];
      }
    }
    answer[i] = sum;
  }
  return answer;
}
```