user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x414(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L104
L114:
  lea    -0x408(%rbp),%rdx
  lea    -0x400(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x408(%rbp),%rax
  test   %rax,%rax
  jne    L105
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L104
L105:
  lea    -0x400(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cltq
  movzbl -0x400(%rbp,%rax,1),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,-0x40c(%rbp)
  cmpl   $0x0,-0x40c(%rbp)
  je     L4
  cmpl   $0x1,-0x40c(%rbp)
  je     L4
  cmpl   $0x5,-0x40c(%rbp)
  je     L4
  cmpl   $0x6,-0x40c(%rbp)
  jne    L32
L4:
  mov    -0x40c(%rbp),%eax
  mov    %eax,-0x410(%rbp)
  jmp    L130
L32:
  cmpl   $0x2,-0x40c(%rbp)
  je     L7
  cmpl   $0x3,-0x40c(%rbp)
  je     L7
  cmpl   $0x7,-0x40c(%rbp)
  je     L7
  cmpl   $0x8,-0x40c(%rbp)
  jne    L17
L7:
  mov    -0x408(%rbp),%rax
  cqto
  shr    $0x3e,%rdx
  add    %rdx,%rax
  and    $0x3,%eax
  sub    %rdx,%rax
  add    $0x4,%eax
  mov    %eax,%edx
  mov    -0x40c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mov    %eax,-0x410(%rbp)
  jmp    L130
L17:
  cmpl   $0x4,-0x40c(%rbp)
  je     L138
  cmpl   $0x9,-0x40c(%rbp)
  jne    L130
L138:
  mov    -0x408(%rbp),%rax
  cqto
  shr    $0x3f,%rdx
  add    %rdx,%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  add    $0x2,%eax
  mov    %eax,%edx
  mov    -0x40c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mov    %eax,-0x410(%rbp)
L130:
  mov    -0x410(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L104:
  mov    -0x414(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x414(%rbp)
  test   %eax,%eax
  jne    L114
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L149
  call   <__stack_chk_fail@plt>
L149:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_guessed: string
name: disassemble_data
arguments: data_tag: D25
type_guessed: string
name: disassemble_data
arguments: data_tag: D4
type_guessed: string
name: disassemble_data
arguments: data_tag: D22
type_guessed: string

tool: D2: %d

tool: D25: %s %lld

tool: D4: 1

tool: D22: %d


assistant: ```c
int main() {
  int i, j, c, len, num, temp, t, res;
  long long int b;
  char a[1001];
  scanf("%d", &t);
  while (t--) {
    scanf("%s %lld", &a, &b);
    if (b == 0) {
      printf("1\n");
      continue;
    }
    c = a[strlen(a) - 1] - '0';
    if (c == 0 || c == 1 || c == 5 || c == 6)
      res = c;
    else if (c == 2 || c == 3 || c == 7 || c == 8)
      res = pow(c, 4 + (b % 4));
    else if (c == 4 || c == 9)
      res = pow(c, 2 + (b % 2));
    printf("%d\n", res % 10);
  }
  return 0;
}
```