user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0xa,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x288,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x278(%rsp)
  xor    %eax,%eax
  call   <putchar@plt>
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%r9
  test   %r9,%r9
  jle    L27
  mov    %r9,%rcx
  xor    %esi,%esi
  lea    0x10(%rsp),%r10
  movabs $0xcccccccccccccccd,%r8
  nop
L41:
  mov    %rcx,%rax
  mov    %rcx,%rdi
  mov    %rsi,%rbp
  mul    %r8
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rdi
  mov    %rcx,%rax
  mov    %rdx,%rcx
  mov    %edi,(%r10,%rsi,4)
  lea    0x1(%rsi),%rsi
  cmp    $0x9,%rax
  jg     L41
  lea    0x18(%rsp,%rbp,4),%r8
  xor    %r11d,%r11d
  xor    %edi,%edi
  nopl   0x0(%rax,%rax,1)
L77:
  mov    (%r10,%r11,4),%ecx
  mov    %r10,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L78:
  xor    %esi,%esi
  cmp    (%rax),%ecx
  sete   %sil
  add    $0x4,%rax
  add    %esi,%edx
  cmp    %r8,%rax
  jne    L78
  cmp    $0x1,%edx
  jg     L133
L115:
  lea    0x1(%r11),%rax
  cmp    %rbp,%r11
  je     L58
L138:
  mov    %rax,%r11
  jmp    L77
  cs nopw 0x0(%rax,%rax,1)
L133:
  movslq %edi,%rbx
  add    $0x1,%edi
  mov    0xe0(%rsp,%rbx,4),%esi
  movslq %edi,%rax
  lea    0x2(%r11,%rbx,1),%rbx
  jmp    L86
  nopw   0x0(%rax,%rax,1)
L53:
  mov    %ecx,%esi
L86:
  lea    -0x1(%rax),%edi
  cmp    %esi,%ecx
  je     L115
  mov    %ecx,0xe0(%rsp,%rax,4)
  mov    %eax,%edi
  mov    %edx,0x1b0(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L53
  lea    0x1(%r11),%rax
  cmp    %rbp,%r11
  jne    L138
  nopl   0x0(%rax,%rax,1)
L58:
  test   %edi,%edi
  je     L27
  movslq %edi,%rbx
  lea    0xe0(%rsp),%r13
  lea    0x1b0(%rsp),%r12
  lea    D4(%rip),%rbp
  jmp    L76
L14:
  mov    0x8(%rsp),%r9
L76:
  mov    (%r12,%rbx,4),%edx
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r9,%rcx
  mov    %rbp,%rdi
  xor    %eax,%eax
  sub    $0x1,%rbx
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L14
L96:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x278(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L129
  add    $0x288,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L27:
  mov    %r9,%rsi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L96
L129:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D0
data_type: string
name: disassemble_data
arguments: data_label: D22
data_type: string
name: disassemble_data
arguments: data_label: D4
data_type: string
name: disassemble_data
arguments: data_label: D11
data_type: string

tool: D0:   Enter Number: 

tool: D22: %lli

tool: D4:   %d is Met %d Times in the Number %lli 


tool: D11:   Everything in the Number %lli is Met Only Once 


assistant: ```c
int main() {
  int a[50], b[50], c[50], d[50], m = 0, i, t, u, v = 0, o;
  long long int n, p;
  printf("\n");
  printf("  Enter Number: ");
  scanf("%lli", &n);
  p = n;
  for (i = 0; p > 0; i++) {
    a[i] = p % 10;
    p = p / 10;
  }
  for (t = 0; t < i; t++) {
    for (u = 0; u <= i; u++) {
      if (a[t] == a[u])
        m++;
    }
    b[t] = m;
    m = 0;
    if (b[t] > 1) {
      o = t;
      while (o >= 0) {
        if (a[t] == c[v])
          break;
        else {
          v++;
          c[v] = a[t];
          d[v] = b[t];
        }
        o--;
      }
    }
  }
  if (v > 0) {
    for (; v > 0; v--) {
      printf("  %d is Met %d Times in the Number %lli \n", c[v], d[v], n);
    }
  } else
    printf("  Everything in the Number %lli is Met Only Once \n", n);
  printf("\n");
  return 0;
}
```