user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D0(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x2c(%rsp),%rsi
  call   <scanf@plt>
  mov    0x2c(%rsp),%edx
  test   %edx,%edx
  jle    L7
  lea    0x34(%rsp),%rax
  movl   $0x0,0x8(%rsp)
  lea    D39(%rip),%rbp
  mov    %rax,0x18(%rsp)
  lea    0x30(%rsp),%rax
  mov    %rax,0x10(%rsp)
  jmp    L30
  nopl   0x0(%rax)
L207:
  mov    0x30(%rsp),%eax
  cmp    $0x1,%eax
  je     L103
  cmp    $0x1,%ebx
  je     L181
L49:
  lea    (%rbx,%rbx,1),%ecx
  cltd
  idiv   %ecx
  mov    %eax,0xc(%rsp)
  mov    %edx,%r13d
  test   %edx,%edx
  je     L123
  lea    D40(%rip),%rdi
  call   <puts@plt>
L197:
  addl   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%eax
  cmp    %eax,0x2c(%rsp)
  jle    L7
L30:
  mov    0x18(%rsp),%rdx
  mov    0x10(%rsp),%rsi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x34(%rsp),%ebx
  test   %ebx,%ebx
  jne    L207
  mov    0x30(%rsp),%eax
  test   %eax,%eax
  jle    L208
  cs nopw 0x0(%rax,%rax,1)
L62:
  add    $0x1,%ebx
  xor    %eax,%eax
  mov    %rbp,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    0x30(%rsp),%ebx
  jl     L62
L208:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%eax
  cmp    %eax,0x2c(%rsp)
  jg     L30
L7:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L166
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L123:
  test   %eax,%eax
  jle    L208
  nopl   0x0(%rax)
L203:
  mov    %ebx,%r14d
  imul   %r13d,%r14d
  add    $0x1,%r13d
  lea    (%r14,%r14,1),%r12d
  add    %ebx,%r14d
  add    %r14d,%r14d
  add    %r12d,%ebx
  mov    %ebx,%r15d
  cmp    %ebx,%r14d
  jle    L176
L13:
  add    $0x1,%r15d
  xor    %eax,%eax
  mov    %rbp,%rdi
  mov    %r15d,%esi
  call   <printf@plt>
  cmp    %r14d,%r15d
  jne    L13
L176:
  cmp    %ebx,%r12d
  jge    L185
  nopl   0x0(%rax,%rax,1)
L50:
  add    $0x1,%r12d
  xor    %eax,%eax
  mov    %rbp,%rdi
  mov    %r12d,%esi
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L50
L185:
  cmp    0xc(%rsp),%r13d
  je     L208
  mov    0x34(%rsp),%ebx
  jmp    L203
  nopw   0x0(%rax,%rax,1)
L181:
  test   $0x1,%al
  jne    L49
  sub    $0x1,%eax
  jle    L208
  mov    $0x2,%ebx
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L129:
  lea    -0x1(%rbx),%edx
  mov    %ebx,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x30(%rsp),%edx
  add    $0x1,%r12d
  add    $0x2,%ebx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cmp    %eax,%r12d
  jl     L129
  jmp    L208
L103:
  lea    D20(%rip),%rdi
  call   <puts@plt>
  jmp    L197
L166:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D0
type_infered: string
name: parse_data
arguments: label: D39
type_infered: string
name: parse_data
arguments: label: D40
type_infered: string
name: parse_data
arguments: label: D10
type_infered: string
name: parse_data
arguments: label: D11
type_infered: string
name: parse_data
arguments: label: D20
type_infered: string

tool: D0: %d

tool: D39: %d 

tool: D40: -1

tool: D10: %d %d

tool: D11: %d %d 

tool: D20: 1

assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int k;
    scanf("%d %d", &n, &k);
    if (k == 0) {
      for (int i = 0; i < n; i++) {
        printf("%d ", i + 1);
      }
      printf("\n");
    } else {
      if (n == 1) {
        printf("1\n");
      } else {
        if (k == 1 && n % 2 == 0) {
          for (int i = 0; i < n / 2; i++) {
            printf("%d %d ", 2 * i + 2, 2 * i + 1);
          }
          printf("\n");
        } else if (n % (2 * k) == 0) {
          int partitions = n / (2 * k);
          for (int i = 0; i < partitions; i++) {
            int start = 2 * k * i;
            int end = 2 * k * (i + 1);
            int mid = start + k;
            for (int j = mid; j < end; j++) {
              printf("%d ", j + 1);
            }
            for (int j = start; j < mid; j++) {
              printf("%d ", j + 1);
            }
          }
          printf("\n");
        } else {
          printf("-1\n");
        }
      }
    }
  }
  return 0;
}
```