user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  lea    D29(%rip),%rbx
  jmp    L207
L121:
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
L207:
  mov    0xc(%rsp),%edi
  call   <rightInt@plt>
  test   %eax,%eax
  je     L121
  mov    0xc(%rsp),%r12d
  test   %r12d,%r12d
  je     L31
  lea    0x10(%rsp),%rsi
  mov    %rsi,%rcx
  mov    $0x0,%ebx
L145:
  movslq %r12d,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %r12d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%r12d
  mov    %r12d,(%rcx)
  mov    %eax,%r12d
  mov    %ebx,%r8d
  add    $0x1,%ebx
  add    $0x4,%rcx
  test   %eax,%eax
  jne    L145
  mov    %ebx,0xc(%rsp)
  test   %ebx,%ebx
  jle    L31
  movslq %r8d,%rax
  shl    $0x2,%rax
  lea    0x10(%rsp,%rax,1),%r11
  mov    %r12d,%edi
  lea    -0x4(%rsi,%rax,1),%r10
  lea    -0x1(%r8),%r9d
  jmp    L15
L168:
  sub    $0x4,%rax
  cmp    %rsi,%rax
  je     L210
L52:
  mov    -0x4(%rax),%edx
  mov    (%rax),%ecx
  cmp    %ecx,%edx
  jle    L168
  mov    %ecx,-0x4(%rax)
  mov    %edx,(%rax)
  jmp    L168
L210:
  lea    0x1(%rdi),%eax
  cmp    %edi,%r8d
  jle    L152
  mov    %eax,%edi
L15:
  cmp    %edi,%r8d
  jle    L210
  mov    %r9d,%eax
  sub    %edi,%eax
  shl    $0x2,%rax
  mov    %r10,%rsi
  sub    %rax,%rsi
  mov    %r11,%rax
  jmp    L52
L152:
  mov    %r12d,%eax
  lea    0x10(%rsp),%r13
  mov    $0x1,%ebp
  lea    D2(%rip),%r14
  jmp    L130
L88:
  test   %ecx,%ecx
  jne    L98
L219:
  mov    0xc(%rsp),%ebx
  cmp    %eax,%ebx
  jle    L31
L130:
  lea    0x1(%rax),%edi
  cmp    %ebx,%edi
  jge    L118
  cltq
  mov    0x10(%rsp,%rax,4),%esi
  lea    0x0(%r13,%rax,4),%rdx
  mov    %r12d,%ecx
  mov    %edi,%eax
L159:
  cmp    0x4(%rdx),%esi
  jne    L88
  add    $0x1,%eax
  add    $0x4,%rdx
  mov    %ebp,%ecx
  cmp    %ebx,%eax
  jne    L159
L25:
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  jmp    L219
L98:
  mov    %eax,%ebx
  jmp    L25
L118:
  mov    %edi,%eax
  jmp    L219
L31:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D29
data_type: string
name: extract_symbolic_data
arguments: data_tag: D2
data_type: string

tool: D29: %d

tool: D2: %d 


assistant: ```c
int main() {
  int i, j, n, a, digits[10];
  scanf("%d", &n);
  while (!rightInt(n)) {
    scanf("%d", &n);
  }
  for (i = 0; n != 0; i++) {
    digits[i] = n % 10;
    n /= 10;
  }
  n = i;
  for (i = 0; i < n; i++) {
    for (j = n - 1; j > i; j--) {
      if (digits[j - 1] > digits[j]) {
        a = digits[j - 1];
        digits[j - 1] = digits[j];
        digits[j] = a;
      }
    }
  }
  for (i = 0, a = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (digits[i] == digits[j]) {
        a = 1;
      } else {
        break;
      }
    }
    if (a) {
      printf("%d \n", digits[i]);
    }
    i = j - 1;
    a = 0;
  }
  return 0;
}
```