user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D65(%rip),%rsi
  lea    D28(%rip),%rdi
  push   %r12
  lea    D54(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%r13
  call   <fopen@plt>
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rax,%rdi
  lea    0x20(%rsp),%r8
  lea    D64(%rip),%rsi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <fscanf@plt>
  mov    0x8(%rsp),%rax
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  cmpq   $0x0,0x20(%rsp)
  mov    %rax,%rbx
  je     L6
  xchg   %ax,%ax
L5:
  xor    %eax,%eax
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fscanf@plt>
  mov    0x18(%rsp),%rax
  movb   $0x1,(%rbx,%rax,1)
  subq   $0x1,0x20(%rsp)
  jne    L5
L6:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x10(%rsp),%rdi
  add    0x8(%rsp),%rdi
  mov    $0x8,%esi
  add    $0x1,%rdi
  call   <calloc@plt>
  mov    0x8(%rsp),%r8
  mov    %rax,%rdi
  mov    0x10(%rsp),%rax
  lea    (%rdi,%rax,8),%rbp
  movq   $0x1,0x0(%rbp)
  movq   $0x0,0x18(%rsp)
  test   %r8,%r8
  jle    L62
  mov    $0x1,%esi
  mov    $0x1,%ecx
  movabs $0x89705f3112a28fe5,%r9
  jmp    L25
  xchg   %ax,%ax
L66:
  mov    0x0(%rbp,%rsi,8),%rcx
  add    $0x1,%rsi
L25:
  add    0x20(%rsp),%rcx
  sub    -0x8(%rdi,%rsi,8),%rcx
  add    $0x3b9aca07,%rcx
  mov    %rcx,%rax
  imul   %r9
  mov    %rcx,%rax
  sar    $0x3f,%rax
  add    %rcx,%rdx
  sar    $0x1d,%rdx
  sub    %rax,%rdx
  imul   $0x3b9aca07,%rdx,%rax
  sub    %rax,%rcx
  mov    %rcx,0x20(%rsp)
  cmpb   $0x0,(%rbx,%rsi,1)
  jne    L27
  mov    %rcx,0x0(%rbp,%rsi,8)
L27:
  mov    %rsi,0x18(%rsp)
  cmp    %r8,%rsi
  jne    L66
L62:
  lea    D22(%rip),%rsi
  lea    D66(%rip),%rdi
  call   <fopen@plt>
  lea    D9(%rip),%rsi
  mov    %rax,%r12
  mov    0x8(%rsp),%rax
  mov    %r12,%rdi
  mov    0x0(%rbp,%rax,8),%rdx
  call   <fprintf@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D65
type: string
name: retrieve_data
arguments: data_label: D28
type: string
name: retrieve_data
arguments: data_label: D54
type: string
name: retrieve_data
arguments: data_label: D64
type: string
name: retrieve_data
arguments: data_label: D22
type: string
name: retrieve_data
arguments: data_label: D66
type: string
name: retrieve_data
arguments: data_label: D9
type: string

tool: D65: r

tool: D28: quacau.inp

tool: D54: %ld

tool: D64: %ld %ld %Ld

tool: D22: w

tool: D66: quacau.out

tool: D9: %Ld


assistant: ```c
int main() {
  FILE *f = fopen("quacau.inp", "r");
  long n, m, i;
  long long k, *ways;
  char *broken;
  fscanf(f, "%ld %ld %Ld", &n, &m, &k);
  broken = calloc(n + 1, sizeof(char));
  for (; k; k--) {
    fscanf(f, "%ld", &i);
    broken[i] = 1;
  }
  fclose(f);
  ways = (long long *)calloc(n + m + 1, sizeof(long long)) + m;
  *ways = 1;
  for (i = 0; i < n; i++) {
    k = (ways[i] + k - ways[i - m] + 1000000007) % 1000000007;
    if (!broken[i + 1])
      ways[i + 1] = k;
  }
  f = fopen("quacau.out", "w");
  fprintf(f, "%Ld\n", ways[n]);
  fclose(f);
  return 0;
}
```