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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x28(%rbp)
  lea    -0x3c(%rbp),%rdx
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x34(%rbp),%rdx
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x2c(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L78
L3:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x38(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x40(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L31
  mov    -0x3c(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L31
  addl   $0x1,-0x28(%rbp)
L31:
  addl   $0x1,-0x24(%rbp)
L78:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L3
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x28(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L57
L74:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x40(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L61
  mov    -0x3c(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L61
  addl   $0x1,-0x28(%rbp)
L61:
  addl   $0x1,-0x20(%rbp)
L57:
  mov    -0x2c(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jl     L74
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L66
  call   <__stack_chk_fail@plt>
L66:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D51
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D47
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D24
type_infered: string

tool: D51: %d %d

tool: D47: %d

tool: D24: %d


assistant: ```c
int main() {
  int count = 0, x;
  int s;
  int t;
  scanf("%d %d", &s, &t);
  int a;
  int b;
  scanf("%d %d", &a, &b);
  int m;
  int n;
  scanf("%d %d", &m, &n);
  int *apple = malloc(sizeof(int) * m);
  for (int apple_i = 0; apple_i < m; apple_i++) {
    scanf("%d", &apple[apple_i]);
    x = a + apple[apple_i];
    if (x >= s && x <= t)
      count++;
  }
  printf("%d\n", count);
  count = 0;
  int *orange = malloc(sizeof(int) * n);
  for (int orange_i = 0; orange_i < n; orange_i++) {
    scanf("%d", &orange[orange_i]);
    x = b + orange[orange_i];
    if (x >= s && x <= t)
      count++;
  }
  printf("%d\n", count);
  return 0;
}
```