user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x34(%rbp)
  movl   $0x1,-0x30(%rbp)
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L42
L81:
  mov    -0x38(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    %dl,-0x20(%rbp,%rax,1)
  mov    -0x34(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x38(%rbp)
  addl   $0x1,-0x34(%rbp)
L42:
  mov    -0x38(%rbp),%eax
  test   %eax,%eax
  jg     L81
  movl   $0x0,-0x2c(%rbp)
  jmp    L23
L98:
  mov    -0x2c(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    $0x63,%al
  je     L59
  movl   $0x0,-0x28(%rbp)
  jmp    L86
L33:
  mov    -0x2c(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  je     L47
  mov    -0x2c(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%edx
  mov    -0x28(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    %al,%dl
  jne    L47
  addl   $0x1,-0x30(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  movb   $0x63,-0x20(%rbp,%rax,1)
L47:
  addl   $0x1,-0x28(%rbp)
L86:
  mov    -0x28(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L33
  mov    -0x2c(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    -0x30(%rbp),%edx
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x30(%rbp)
  jmp    L31
L59:
  nop
L31:
  addl   $0x1,-0x2c(%rbp)
L23:
  mov    -0x2c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L98
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L97
  call   <__stack_chk_fail@plt>
L97:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D32
type_infered: string
name: retrieve_data
arguments: tag: D27
type_infered: string
name: retrieve_data
arguments: tag: D38
type_infered: string
name: retrieve_data
arguments: tag: D29
type_infered: string

tool: D32: Input the number :

tool: D27: %d

tool: D38: %d


tool: D29: %d - %d


assistant: ```c
int main() {
  char arr[20];
  int modnum;
  int number;
  int i = 0, counter = 1, j, k;
  printf("Input the number :");
  scanf("%d", &number);
  while (number > 0) {
    modnum = number % 10;
    arr[i] = modnum;
    printf("%d\n", arr[i]);
    number /= 10;
    i++;
  }
  for (j = 0; j < i; j++) {
    if (arr[j] == 'c')
      continue;
    for (k = 0; k < i; k++) {
      if (j != k) {
        if (arr[j] == arr[k]) {
          counter++;
          arr[k] = 'c';
        }
      }
    }
    printf("%d - %d\n", arr[j], counter);
    counter = 1;
  }
}
```