user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D14(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x698,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x688(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rax
  movl   $0x1,0x1c(%rsp)
  lea    0x30(%rsp),%r13
  mov    %rax,0x10(%rsp)
  lea    0x2c(%rsp),%rax
  mov    %rax,0x8(%rsp)
L27:
  mov    0x10(%rsp),%rdi
  mov    $0x640,%edx
  xor    %esi,%esi
  xor    %r14d,%r14d
  call   <memset@plt>
  nopl   0x0(%rax)
L62:
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jle    L65
  mov    0x2c(%rsp),%edx
  test   %edx,%edx
  jle    L14
  movslq %r14d,%rbp
  xor    %r15d,%r15d
  lea    0x0(%rbp,%rbp,4),%rbx
  shl    $0x2,%rbx
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    0x30(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x30(%rsp)
  cltq
  lea    (%rax,%rax,4),%rcx
  add    %rbx,%rax
  lea    0x0(%rbp,%rcx,4),%rcx
  movl   $0x1,0x40(%rsp,%rcx,4)
  movl   $0x1,0x40(%rsp,%rax,4)
  cmp    %r15d,0x2c(%rsp)
  jg     L39
L14:
  add    $0x1,%r14d
  cmp    $0x13,%r14d
  jne    L62
  mov    0x10(%rsp),%rdi
  mov    $0x14,%esi
  call   <computeAllPairShortestPath@plt>
  mov    0x1c(%rsp),%esi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x34(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x34(%rsp),%eax
  test   %eax,%eax
  jle    L43
  xor    %ebx,%ebx
  lea    0x3c(%rsp),%r14
  lea    0x38(%rsp),%rbp
  lea    D31(%rip),%r15
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r14,%rdx
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x38(%rsp),%esi
  mov    %r15,%rdi
  add    $0x1,%ebx
  mov    0x3c(%rsp),%r8d
  lea    -0x1(%rsi),%eax
  lea    -0x1(%r8),%edx
  cltq
  movslq %edx,%rdx
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,4),%rax
  mov    %r8d,%edx
  mov    0x40(%rsp,%rax,4),%ecx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %ebx,0x34(%rsp)
  jg     L0
L43:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0x1c(%rsp)
  jmp    L27
L65:
  mov    0x688(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x698,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D14
datatype: string
name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D31
datatype: string
name: unpack_data
arguments: tag: D8
datatype: string

tool: D14: %d

tool: D16: Test Set #%d


tool: D31: %2d to %2d: %d


tool: D8: %d %d

assistant: ```c
int main(void) {
  int caseId, adjMat[20][20], numNeighbor, neighbor, stopFlg, city, time, numQuery, cityOne, cityTwo, query;
  caseId = 1;
  stopFlg = 0;
  while (1) {
    memset(adjMat, 0, 20 * 20 * sizeof(int));
    for (city = 0; city < 20 - 1; city++)
      if (scanf("%d", &numNeighbor) > 0) {
        for (time = 0; time < numNeighbor; time++) {
          scanf("%d", &neighbor);
          neighbor--;
          adjMat[neighbor][city] = 1;
          adjMat[city][neighbor] = 1;
        }
      } else {
        stopFlg = 1;
        break;
      }
    if (stopFlg)
      break;
    computeAllPairShortestPath(adjMat, 20);
    printf("Test Set #%d\n", caseId);
    scanf("%d", &numQuery);
    for (query = 0; query < numQuery; query++) {
      scanf("%d %d", &cityOne, &cityTwo);
      printf("%2d to %2d: %d\n", cityOne, cityTwo, adjMat[cityOne - 1][cityTwo - 1]);
    }
    printf("\n");
    caseId++;
  }
  return 0;
}
```