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

```asm
<main>:
  endbr64
  push   %r15
  xor    %r10d,%r10d
  xor    %r8d,%r8d
  xor    %r9d,%r9d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x188,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x178(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L9:
  mov    0x8(%rsp),%rax
  lea    0x6(%r8),%esi
  add    $0x24,%r8d
  lea    (%rax,%r10,1),%rdi
L105:
  lea    -0x6(%rsi),%eax
  mov    %rdi,%rdx
  nopw   0x0(%rax,%rax,1)
L115:
  mov    %eax,%ecx
  add    $0x1,%eax
  add    $0x4,%rdx
  mov    %ecx,-0x4(%rdx)
  cmp    %esi,%eax
  jne    L115
  lea    0x6(%rax),%esi
  add    $0x18,%rdi
  cmp    %r8d,%esi
  jne    L105
  add    $0x1,%r9d
  cmp    $0x2,%r9d
  je     L40
  cmp    $0x3,%r9d
  je     L126
  mov    $0x64,%r8d
L145:
  add    $0x78,%r10
  jmp    L9
L126:
  lea    D46(%rip),%rdi
  xor    %r15d,%r15d
  lea    D19(%rip),%rbp
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  lea    0x78(%rax),%r12
  mov    %r12,%rbx
L1:
  mov    %r15d,%esi
  lea    D5(%rip),%rdi
  lea    -0x78(%rbx),%r13
  xor    %eax,%eax
  call   <printf@plt>
L48:
  xor    %r14d,%r14d
  xchg   %ax,%ax
L140:
  mov    0x0(%r13,%r14,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <printf@plt>
  cmp    $0x6,%r14
  jne    L140
  mov    $0xa,%edi
  add    $0x18,%r13
  call   <putchar@plt>
  cmp    %rbx,%r13
  jne    L48
  add    $0x1,%r15d
  lea    0x78(%r13),%rbx
  cmp    $0x3,%r15d
  jne    L1
  mov    0x8(%rsp),%r13
  mov    %r12,%rsi
  add    $0x1e0,%r13
L89:
  lea    -0x78(%rsi),%rax
L127:
  mov    (%rax),%edx
  mov    0x14(%rax),%ecx
  add    $0x18,%rax
  mov    %ecx,-0x18(%rax)
  mov    -0x8(%rax),%ecx
  mov    %edx,-0x4(%rax)
  mov    -0x14(%rax),%edx
  mov    %ecx,-0x14(%rax)
  mov    -0xc(%rax),%ecx
  mov    %edx,-0x8(%rax)
  mov    -0x10(%rax),%edx
  mov    %ecx,-0x10(%rax)
  mov    %edx,-0xc(%rax)
  cmp    %rsi,%rax
  jne    L127
  lea    0x78(%rax),%rsi
  cmp    %r13,%rsi
  jne    L89
  lea    D15(%rip),%rdi
  xor    %r13d,%r13d
  call   <puts@plt>
L43:
  mov    %r13d,%esi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x78(%r12),%r15
L131:
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L50:
  mov    (%r15,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x6,%rbx
  jne    L50
  mov    $0xa,%edi
  add    $0x18,%r15
  call   <putchar@plt>
  cmp    %r12,%r15
  jne    L131
  add    $0x1,%r13d
  lea    0x78(%r15),%r12
  cmp    $0x3,%r13d
  jne    L43
  mov    0x178(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L113
  add    $0x188,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L40:
  mov    $0x3e8,%r8d
  jmp    L145
L113:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D46
datatype: string
name: disassemble_data
arguments: data_label: D19
datatype: string
name: disassemble_data
arguments: data_label: D5
datatype: string
name: disassemble_data
arguments: data_label: D15
datatype: string

tool: D46: 
Wypelniona tablica reprezentujaca obraz:


tool: D19: %d 

tool: D5: Kanal nr: %d


tool: D15: 
Po odbiciu wzgledem osi pionowej:


assistant: ```c
int main(void) {
  int obraz[3][5][6];
  int i, j, k, l, wartosc = 0;
  int temp;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 5; j++) {
      for (k = 0; k < 6; k++) {
        obraz[i][j][k] = wartosc++;
      }
    }
    wartosc = (i == 1) ? 1000 : 100;
  }
  printf("\nWypelniona tablica reprezentujaca obraz:\n\n");
  for (i = 0; i < 3; i++) {
    printf("Kanal nr: %d\n", i);
    for (j = 0; j < 5; j++) {
      for (k = 0; k < 6; k++) {
        printf("%d ", obraz[i][j][k]);
      }
      printf("\n");
    }
  }
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 5; j++) {
      for (k = 0, l = 6 - 1; k < l; k++, l--) {
        temp = obraz[i][j][k];
        obraz[i][j][k] = obraz[i][j][l];
        obraz[i][j][l] = temp;
      }
    }
  }
  printf("\nPo odbiciu wzgledem osi pionowej:\n\n");
  for (i = 0; i < 3; i++) {
    printf("Kanal nr: %d\n", i);
    for (j = 0; j < 5; j++) {
      for (k = 0; k < 6; k++) {
        printf("%d ", obraz[i][j][k]);
      }
      printf("\n");
    }
  }
  return 0;
}
```