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

```asm
<print_permutation>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  cmp    %edi,%edx
  je     L32
  mov    %edi,%r15d
  test   %edi,%edi
  jle    L37
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    -0x1(%rdx),%r8d
  mov    $0x1,%r12d
  lea    (%rsi,%rax,4),%rax
  and    $0xfffffffc,%ecx
  lea    0x1(%rdx),%r13d
  mov    %rax,0x8(%rsp)
  mov    %edx,%eax
  shr    $0x2,%eax
  sub    $0x1,%eax
  shl    $0x4,%rax
  lea    0x10(%rsi,%rax,1),%r14
L159:
  test   %ebp,%ebp
  jle    L53
  cmp    $0x2,%r8d
  jbe    L102
  movd   %r12d,%xmm3
  movdqa D2(%rip),%xmm1
  mov    %rbx,%rax
  pshufd $0x0,%xmm3,%xmm2
  nopl   0x0(%rax)
L49:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  pcmpeqd %xmm2,%xmm0
  pandn  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  cmp    %r14,%rax
  jne    L49
  movdqa %xmm0,%xmm2
  mov    %ecx,%edx
  psrldq $0x8,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm1
  pandn  %xmm2,%xmm0
  por    %xmm1,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm2,%xmm1
  por    %xmm0,%xmm1
  movd   %xmm1,%eax
  cmp    %ecx,%ebp
  je     L81
L155:
  movslq %edx,%rdi
  add    $0x1,%edx
  lea    0x0(,%rdi,4),%rsi
  cmp    %r12d,(%rbx,%rdi,4)
  je     L162
  cmp    %ebp,%edx
  jge    L81
  add    $0x1,%edx
  cmp    %r12d,0x4(%rbx,%rsi,1)
  je     L141
L68:
  cmp    %edx,%ebp
  jle    L81
L15:
  cmp    0x8(%rbx,%rsi,1),%r12d
  je     L67
L81:
  test   %eax,%eax
  jne    L53
L67:
  add    $0x1,%r12d
  cmp    %r12d,%r15d
  jge    L159
L37:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L53:
  mov    0x8(%rsp),%rax
  mov    %r13d,%edx
  mov    %rbx,%rsi
  mov    %r15d,%edi
  mov    %ecx,0x4(%rsp)
  mov    %r12d,(%rax)
  mov    %r8d,(%rsp)
  call   <print_permutation>
  mov    0x4(%rsp),%ecx
  mov    (%rsp),%r8d
  jmp    L67
L162:
  xor    %eax,%eax
  cmp    %edx,%ebp
  jle    L67
  add    $0x1,%edx
  cmp    %r12d,0x4(%rbx,%rsi,1)
  jne    L68
L141:
  xor    %eax,%eax
  cmp    %ebp,%edx
  jl     L15
  jmp    L67
  nopl   0x0(%rax,%rax,1)
L32:
  test   %edx,%edx
  jle    L5
  lea    -0x1(%rdx),%eax
  lea    D0(%rip),%rbp
  lea    0x4(%rsi,%rax,4),%r12
  nopl   0x0(%rax,%rax,1)
L143:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L143
L5:
  add    $0x18,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <putchar@plt>
L102:
  mov    $0x1,%eax
  xor    %edx,%edx
  jmp    L155
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_guessed: byte[16]
name: disassemble_data
arguments: data_tag: D0
type_guessed: string

tool: {"D2": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D0: %3d

assistant: ```c
void print_permutation(int n, int *a, int cur) {
  int i, j;
  if (cur == n) {
    for (i = 0; i < n; ++i)
      printf("%3d", a[i]);
    printf("\n");
  } else
    for (i = 1; i <= n; ++i) {
      int ok = 1;
      for (j = 0; j < cur; ++j)
        if (a[j] == i)
          ok = 0;
      if (ok) {
        a[cur] = i;
        print_permutation(n, a, cur + 1);
      }
    }
}
```