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

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x708,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x6f8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L36
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  call   <srandom@plt>
L36:
  xor    %ebp,%ebp
  lea    0x30(%rsp),%rbx
  lea    0x270(%rsp),%r12
  nopl   0x0(%rax)
L32:
  xor    %eax,%eax
  call   <random@plt>
  divsd  D5(%rip),%xmm0
  movsd  %xmm0,(%rbx,%rbp,1)
  movsd  %xmm0,(%r12,%rbp,1)
  add    $0x8,%rbp
  cmp    $0x240,%rbp
  jne    L32
  lea    0x4b0(%rsp),%r15
  mov    %rbx,%rsi
  xor    %r13d,%r13d
  mov    %r15,%rdi
  mov    %r15,%r14
  call   <cl_invert@plt>
  mov    %r12,%rdx
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <mul@plt>
  lea    0x38(%rsp),%rax
  mov    %rbx,0x18(%rsp)
  mov    %rax,0x10(%rsp)
  lea    0x4b8(%rsp),%rax
  movq   $0x0,0x28(%rsp)
  mov    %rax,0x8(%rsp)
L71:
  mov    0x28(%rsp),%rax
  xor    %r12d,%r12d
  mov    %r14,%rbp
  mov    %rax,0x20(%rsp)
L8:
  mov    0x20(%rsp),%rax
  xor    %r14d,%r14d
  mov    %rax,(%rsp)
  mov    %rbp,%rax
  mov    %r14d,%ebp
  mov    %rax,%r14
L59:
  mov    (%rsp),%rbx
  xor    %r15d,%r15d
L89:
  mov    0x10(%rsp),%rax
  movsd  (%r14,%rbx,1),%xmm0
  mov    %r15d,%r8d
  mov    %ebp,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  lea    D1(%rip),%rdi
  movsd  (%rax,%rbx,1),%xmm3
  mov    0x18(%rsp),%rax
  movsd  (%rax,%rbx,1),%xmm2
  mov    0x8(%rsp),%rax
  movsd  (%rax,%rbx,1),%xmm1
  add    $0x10,%rbx
  call   <printf@plt>
  cmp    $0x1,%r15d
  jne    L69
  add    $0x1,%ebp
  addq   $0x20,(%rsp)
  cmp    $0x3,%ebp
  jne    L59
  addq   $0x60,0x20(%rsp)
  mov    %r14,%rbp
  cmp    $0x1,%r12d
  je     L49
  mov    $0x1,%r12d
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L69:
  mov    $0x1,%r15d
  jmp    L89
  nopl   0x0(%rax,%rax,1)
L49:
  addq   $0xc0,0x28(%rsp)
  add    $0x1,%r13d
  cmp    $0x3,%r13d
  jne    L71
  mov    0x6f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x708,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D5
type: byte[8]
name: retrieve_data
arguments: tag: D1
type: string

tool: {"D5": [0x00,0x00,0xc0,0xff,0xff,0xff,0xdf,0x41]}

tool: D1:  [%d,%d][%d,%d]  %15f %15f   %15f %15f


assistant: ```c
int main(int argc, char *argv[]) {
  double cl[(2 * 2 * 3 * 3 * 2)];
  double v[(2 * 2 * 3 * 3 * 2)];
  double ci[(2 * 2 * 3 * 3 * 2)];
  int a, i, b, j;
  if (argc > 1)
    srandom(atoi(argv[1]));
  for (i = 0; i < (2 * 2 * 3 * 3 * 2); i++)
    v[i] = cl[i] = random() / (double)0x7fffffff;
  cl_invert(ci, cl);
  mul(cl, ci, v);
  for (a = 0; a < 3; a++)
    for (i = 0; i < 2; i++) {
      for (b = 0; b < 3; b++)
        for (j = 0; j < 2; j++) {
          printf(" [%d,%d][%d,%d]  %15f %15f   %15f %15f\n", a, i, b, j, (&((ci)[2 * ((j) + 2 * ((b) + 3 * ((i) + 2 * (a))))]))[0], (&((ci)[2 * ((j) + 2 * ((b) + 3 * ((i) + 2 * (a))))]))[1], (&((cl)[2 * ((j) + 2 * ((b) + 3 * ((i) + 2 * (a))))]))[0], (&((cl)[2 * ((j) + 2 * ((b) + 3 * ((i) + 2 * (a))))]))[1]);
        }
    }
  return 0;
}
```