user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    D36(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x10(%rsp)
  jle    L91
  mov    $0x0,%r15d
  lea    0x14(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    D36(%rip),%r12
  jmp    L132
L142:
  lea    D28(%rip),%rdi
  call   <printf@plt>
L74:
  cmp    %ebx,0x14(%rsp)
  je     L4
L128:
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  add    $0x1,%r15d
  cmp    %r15d,0x10(%rsp)
  jle    L91
L132:
  mov    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    0x14(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  cmpl   $0x0,0x14(%rsp)
  jle    L51
  mov    %r14,%rbp
  mov    $0x0,%ebx
L48:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x14(%rsp),%eax
  add    $0x4,%rbp
  cmp    %ebx,%eax
  jg     L48
  test   %eax,%eax
  jle    L51
  mov    %r13,%rbp
  mov    $0x0,%ebx
L23:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x14(%rsp),%r8d
  add    $0x4,%rbp
  cmp    %ebx,%r8d
  jg     L23
  test   %r8d,%r8d
  jle    L42
  mov    %r13,%rdx
  mov    $0xffffffff,%esi
  mov    $0x0,%ebx
  mov    %r14,%r9
  sub    %r13,%r9
  jmp    L150
L53:
  mov    %edi,%esi
L46:
  add    $0x1,%ebx
  add    $0x4,%rdx
  cmp    %ebx,%r8d
  je     L4
L150:
  mov    (%rdx),%eax
  mov    (%r9,%rdx,1),%ecx
  cmp    %ecx,%eax
  mov    %ecx,%edi
  cmovle %eax,%edi
  cmp    %esi,%edi
  jge    L53
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  cmp    %esi,%eax
  jl     L142
  mov    %eax,%esi
  jmp    L46
L42:
  mov    $0x0,%ebx
  jmp    L74
L4:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L128
L91:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L111
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  mov    $0x0,%ebx
  jmp    L74
L111:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D36
type_guessed: string
name: extract_symbolic_data
arguments: label: D28
type_guessed: string
name: extract_symbolic_data
arguments: label: D3
type_guessed: string

tool: D36: %d

tool: D28: NO


tool: D3: YES


assistant: ```c
int main() {
  int i, j, T, N, A, B, min, max, last, *seqA, *seqB;
  scanf("%d", &T);
  for (i = 0; i < T; i++) {
    scanf("%d", &N);
    seqA = (int *)malloc(sizeof(int) * N);
    seqB = (int *)malloc(sizeof(int) * N);
    for (j = 0; j < N; j++)
      scanf("%d", &seqA[j]);
    for (j = 0; j < N; j++)
      scanf("%d", &seqB[j]);
    for (j = 0, last = -1; j < N; j++) {
      min = ((seqA[j]) < (seqB[j]) ? (seqA[j]) : (seqB[j]));
      max = ((seqA[j]) > (seqB[j]) ? (seqA[j]) : (seqB[j]));
      if (min >= last)
        last = min;
      else if (max >= last)
        last = max;
      else {
        printf("NO\n");
        break;
      }
    }
    if (j == N)
      printf("YES\n");
    free(seqA);
    free(seqB);
  }
  return 0;
}
```