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

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x64,%ecx
  lea    D14(%rip),%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x678,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x668(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r13
  lea    0x10(%rsp),%rsi
  movq   $0x0,0x10(%rsp)
  mov    %r13,%rdi
  lea    0x340(%rsp),%r12
  movq   $0x0,0x18(%rsp)
  rep stos %rax,%es:(%rdi)
  mov    $0x64,%ecx
  mov    %r12,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %r15,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  cmpq   $0x0,0x10(%rsp)
  mov    %rax,(%rsp)
  je     L57
  mov    %rax,%r14
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L91:
  mov    %r14,%rsi
  mov    %r15,%rdi
  add    $0x1,%rbp
  add    $0x8,%r14
  call   <scanf@plt>
  cmp    %rbp,0x10(%rsp)
  ja     L91
L57:
  lea    0x18(%rsp),%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  mov    0x18(%rsp),%eax
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  mov    0x18(%rsp),%rcx
  mov    %rax,0x8(%rsp)
  test   %rcx,%rcx
  je     L15
  mov    %rax,%r14
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  nopl   0x0(%rax)
L77:
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  mov    (%r14),%rax
  mov    0x18(%rsp),%rcx
  cmp    %rax,%rbp
  cmovb  %rax,%rbp
  add    $0x1,%rbx
  add    $0x8,%r14
  cmp    %rbx,%rcx
  ja     L77
  mov    0x10(%rsp),%rsi
  test   %rsi,%rsi
  je     L9
L24:
  mov    (%rsp),%rax
  lea    (%rax,%rsi,8),%rdi
  nopl   (%rax)
L10:
  mov    %rbp,%rdx
  add    $0x8,%rax
  sub    -0x8(%rax),%rdx
  addq   $0x1,0x20(%rsp,%rdx,8)
  cmp    %rax,%rdi
  jne    L10
  test   %rcx,%rcx
  je     L92
L59:
  mov    0x8(%rsp),%r14
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L48:
  mov    %rbp,%rdx
  add    $0x1,%rax
  sub    (%r14),%rdx
  add    $0x8,%r14
  addq   $0x1,0x340(%rsp,%rdx,8)
  cmp    %rax,%rcx
  ja     L48
  test   %rsi,%rsi
  je     L122
L92:
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L40:
  mov    (%r12,%rax,8),%rdx
  sub    0x0(%r13,%rax,8),%rdx
  mov    %rdx,(%r12,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rsi
  ja     L40
L122:
  lea    0x658(%rsp),%rbx
  sub    $0x63,%rbp
  lea    D5(%rip),%r12
  nopl   0x0(%rax)
L90:
  cmpq   $0x0,(%rbx)
  je     L62
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L62:
  sub    $0x8,%rbx
  add    $0x1,%rbp
  jmp    L90
L15:
  mov    0x10(%rsp),%rbp
  test   %rbp,%rbp
  je     L122
  mov    %rbp,%rsi
  xor    %ebp,%ebp
  jmp    L24
L9:
  test   %rcx,%rcx
  jne    L59
  jmp    L122
```

tool_calls: name: retrieve_data
arguments: data_label: D14
datatype: string
name: retrieve_data
arguments: data_label: D5
datatype: string

tool: D14: %llu

tool: D5: %llu 

assistant: ```c
int main() {
  unsigned long long n = 0, m = 0, max = 0, min = 0, i;
  unsigned long long *arrn, *arrm;
  unsigned long long addn[100] = {0}, addm[100] = {0};
  scanf("%llu", &n);
  arrn = (unsigned long long *)malloc(sizeof(unsigned long long) * n);
  for (i = 0; i < n; i++) {
    scanf("%llu", (arrn + i));
  }
  scanf("%llu", &m);
  arrm = (unsigned long long *)malloc(sizeof(unsigned long long) * m);
  for (i = 0; i < m; i++) {
    scanf("%llu", (arrm + i));
    if (arrm[i] > max) {
      max = arrm[i];
    }
  }
  for (i = 0; i < n; i++) {
    addn[max - arrn[i]] += 1;
  }
  for (i = 0; i < m; i++) {
    addm[max - arrm[i]] += 1;
  }
  for (i = 0; i < n; i++) {
    addm[i] = (addm[i] - addn[i]);
  }
  for (i = 99; i >= 0; i--) {
    if (addm[i] > 0) {
      printf("%llu ", max - i);
    }
  }
  printf("\b");
  return 0;
}
```