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

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  pxor   %xmm11,%xmm11
  movdqa D56(%rip),%xmm7
  xor    %eax,%eax
  movdqa %xmm11,%xmm1
  movdqa %xmm11,%xmm12
  movdqa D55(%rip),%xmm9
  movdqa D31(%rip),%xmm8
  movdqa D52(%rip),%xmm10
  movaps %xmm11,(%rsp)
  movdqa %xmm11,%xmm13
  pcmpgtd %xmm9,%xmm1
  movaps %xmm1,0x20(%rsp)
  movdqa %xmm11,%xmm1
  pcmpgtd %xmm8,%xmm1
  movaps %xmm1,0x10(%rsp)
  movdqa %xmm11,%xmm1
  pcmpgtd %xmm7,%xmm1
  movaps %xmm1,0x30(%rsp)
  nopl   (%rax)
L6:
  movdqa %xmm10,%xmm1
  movdqa %xmm13,%xmm4
  movdqa 0x20(%rsp),%xmm14
  add    $0x1,%eax
  paddd  D54(%rip),%xmm10
  movdqa %xmm1,%xmm0
  punpckldq %xmm1,%xmm0
  movdqa %xmm14,%xmm3
  movdqa %xmm14,%xmm15
  pcmpgtd %xmm0,%xmm4
  pmuludq %xmm0,%xmm3
  movdqa %xmm0,%xmm6
  pmuludq %xmm9,%xmm6
  movdqa %xmm4,%xmm2
  pmuludq %xmm9,%xmm2
  paddq  %xmm3,%xmm2
  movdqa %xmm13,%xmm3
  psllq  $0x20,%xmm2
  paddq  %xmm2,%xmm6
  movdqa %xmm1,%xmm2
  punpckhdq %xmm1,%xmm2
  pcmpgtd %xmm2,%xmm3
  pmuludq %xmm2,%xmm15
  movdqa %xmm2,%xmm14
  pmuludq %xmm9,%xmm14
  movdqa %xmm3,%xmm5
  pmuludq %xmm9,%xmm5
  paddq  %xmm15,%xmm5
  movdqa 0x10(%rsp),%xmm15
  psllq  $0x20,%xmm5
  paddq  %xmm14,%xmm5
  movdqa 0x10(%rsp),%xmm14
  pmuludq %xmm2,%xmm15
  shufps $0xdd,%xmm5,%xmm6
  movdqa %xmm6,%xmm5
  pslld  $0x1,%xmm5
  pmuludq %xmm0,%xmm14
  paddd  %xmm6,%xmm5
  movdqa %xmm4,%xmm6
  pcmpeqd %xmm1,%xmm5
  pmuludq %xmm8,%xmm6
  pmuludq %xmm7,%xmm4
  pand   %xmm1,%xmm5
  paddd  (%rsp),%xmm5
  movaps %xmm5,(%rsp)
  movdqa %xmm0,%xmm5
  paddq  %xmm14,%xmm6
  movdqa %xmm2,%xmm14
  pmuludq %xmm8,%xmm5
  psllq  $0x20,%xmm6
  pmuludq %xmm8,%xmm14
  paddq  %xmm6,%xmm5
  movdqa %xmm3,%xmm6
  pmuludq %xmm8,%xmm6
  pmuludq %xmm7,%xmm3
  paddq  %xmm15,%xmm6
  psllq  $0x20,%xmm6
  paddq  %xmm14,%xmm6
  shufps $0xdd,%xmm6,%xmm5
  psrad  $0x1,%xmm5
  movdqa %xmm5,%xmm6
  pslld  $0x2,%xmm6
  paddd  %xmm5,%xmm6
  pcmpeqd %xmm1,%xmm6
  pand   %xmm1,%xmm6
  paddd  %xmm6,%xmm12
  movdqa 0x30(%rsp),%xmm6
  movdqa %xmm6,%xmm5
  pmuludq %xmm0,%xmm5
  pmuludq %xmm7,%xmm0
  paddq  %xmm5,%xmm4
  psllq  $0x20,%xmm4
  paddq  %xmm4,%xmm0
  movdqa %xmm6,%xmm4
  pmuludq %xmm2,%xmm4
  pmuludq %xmm7,%xmm2
  paddq  %xmm4,%xmm3
  psllq  $0x20,%xmm3
  paddq  %xmm3,%xmm2
  shufps $0xdd,%xmm2,%xmm0
  paddd  %xmm1,%xmm0
  psrad  $0x3,%xmm0
  movdqa %xmm0,%xmm2
  pslld  $0x4,%xmm2
  psubd  %xmm0,%xmm2
  movdqa %xmm2,%xmm0
  pcmpeqd %xmm1,%xmm0
  pand   %xmm1,%xmm0
  paddd  %xmm0,%xmm11
  cmp    $0xfa,%eax
  jne    L6
  movdqa %xmm11,%xmm0
  movdqa (%rsp),%xmm7
  lea    D7(%rip),%rdi
  psrldq $0x8,%xmm0
  paddd  %xmm0,%xmm11
  movdqa %xmm12,%xmm0
  psrldq $0x8,%xmm0
  paddd  %xmm0,%xmm12
  movdqa %xmm7,%xmm0
  psrldq $0x8,%xmm0
  paddd  %xmm7,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%esi
  movdqa %xmm12,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm12
  movdqa %xmm11,%xmm0
  psrldq $0x4,%xmm0
  movd   %xmm12,%eax
  paddd  %xmm0,%xmm11
  add    %eax,%esi
  movd   %xmm11,%eax
  sub    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D56
datatype: byte[16]
name: read_data
arguments: data_tag: D55
datatype: byte[16]
name: read_data
arguments: data_tag: D31
datatype: byte[16]
name: read_data
arguments: data_tag: D52
datatype: byte[16]
name: read_data
arguments: data_tag: D54
datatype: byte[16]
name: read_data
arguments: data_tag: D7
datatype: string

tool: {"D56": [0x89,0x88,0x88,0x88,0x89,0x88,0x88,0x88,0x89,0x88,0x88,0x88,0x89,0x88,0x88,0x88]}

tool: {"D55": [0x56,0x55,0x55,0x55,0x56,0x55,0x55,0x55,0x56,0x55,0x55,0x55,0x56,0x55,0x55,0x55]}

tool: {"D31": [0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66]}

tool: {"D52": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D54": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D7: %d


assistant: ```c
int main(void) {
  int i, m1 = 0, m2 = 0, m3 = 0;
  for (i = 0; i < 1000; i++) {
    if (i % 3 == 0) {
      m1 += i;
    }
    if (i % 5 == 0) {
      m2 += i;
    }
    if (i % 15 == 0) {
      m3 += i;
    }
  }
  printf("%d\n", m1 + m2 - m3);
  return 0;
}
```