user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xb000(%rsp),%r11
L50:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L50
  sub    $0xd30,%rsp
  mov    $0x226,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0xbd28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  rep stos %rax,%es:(%rdi)
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L31
  lea    0x1140(%rsp),%r13
  xor    %ebp,%ebp
  lea    D11(%rip),%r14
  mov    %r13,%rbx
L81:
  mov    %rbx,%rsi
  lea    0x14(%rbx),%rdx
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  add    $0x1,%ebp
  add    $0x28,%rbx
  cmp    %ebp,%esi
  jg     L81
  test   %esi,%esi
  jle    L31
  lea    0x1154(%rsp),%rbp
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  nopl   0x0(%rax)
L32:
  mov    %rbp,%rdi
  call   <modify@plt>
  test   %rax,%rax
  je     L77
  movl   $0x1,(%r12,%rbx,4)
  add    $0x1,%r14d
L77:
  mov    0xc(%rsp),%esi
  add    $0x1,%rbx
  add    $0x28,%rbp
  cmp    %ebx,%esi
  jg     L32
  test   %r14d,%r14d
  jne    L36
  cmp    $0x1,%esi
  je     L99
L31:
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L60:
  mov    0xbd28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L103
  add    $0xbd30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L99:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  jmp    L60
L36:
  xor    %eax,%eax
  mov    %r14d,%esi
  lea    D35(%rip),%rdi
  xor    %ebx,%ebx
  call   <printf@plt>
  cmpl   $0x0,0xc(%rsp)
  lea    D22(%rip),%rbp
  jg     L9
  jmp    L60
  nop
L88:
  add    $0x1,%rbx
  add    $0x28,%r13
  cmp    %ebx,0xc(%rsp)
  jle    L60
L9:
  mov    (%r12,%rbx,4),%eax
  test   %eax,%eax
  je     L88
  lea    0x14(%r13),%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L88
L103:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D28
type: string
name: unpack_data
arguments: data_tag: D11
type: string
name: unpack_data
arguments: data_tag: D21
type: string
name: unpack_data
arguments: data_tag: D19
type: string
name: unpack_data
arguments: data_tag: D35
type: string
name: unpack_data
arguments: data_tag: D22
type: string

tool: D28: %d

tool: D11: %s %s

tool: D21: There are %d accounts and no account is modified


tool: D19: There is 1 account and no account is modified

tool: D35: %d


tool: D22: %s %s


assistant: ```c
int main() {
  char password[1100][2][20];
  int flag[1100] = {0};
  int n, count = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%s %s", password[i][0], password[i][1]);
  for (int i = 0; i < n; i++) {
    if (modify(password[i][1])) {
      count++;
      flag[i] = 1;
    }
  }
  if (count == 0) {
    if (n == 1)
      printf("There is 1 account and no account is modified\n");
    else
      printf("There are %d accounts and no account is modified\n", n);
  } else {
    printf("%d\n", count);
    for (int i = 0; i < n; i++) {
      if (flag[i])
        printf("%s %s\n", password[i][0], password[i][1]);
    }
  }
  return 0;
}
```