void pr_binary(int num){
    for(int i=10;i>=0;i--) cout<<((num>>i)&1);
    cout<<endl;
}

//Update :You can also represent any number in its binary form as;
cout<<bitset<const_length>(number);

if((a&(1<<i))!=0) cout<<"set"<<endl;
  // check if set or not;
  else cout<<"Not set"<<endl;

  int ans=0;
  for (int i=31; i>=0;--i)
  {
    if((a&(1<<i))!=0) ans++;
  }
  dis(ans);

//Even though the inbuilt function is also there.

  cout<<__builtin_popcountll((1ll<<35)-1);

  pr_binary(a | (1<<i));
  // set that ith bit;

  pr_binary(a&(~(1<<i)));
  // unset the ith bit;

  pr_binary(a ^ (1<<i));
  // toggle the ith bit from set to unset and vice-versa;

    int n,od=0,ev=0;
    cin>>n;
    for(int i=1;i<=n;i++){
     if(i&1) od++;
     else ev++;
  } 
  cout<<"Count of Odd"<<od<<endl;
  cout<<"Count of Even"<<ev<<endl;

//Although the arithmetic operations are fast ,but by bits manipulation we can make them //more faster.
   int n=5;
   n=n>>1;
  // divide by two
  dis(n);
  n=n<<1;
  // multiply by two

    for(char c='A';c<='Z';c++){
      cout<<c<<" ";
      pr_binary(int(c));
    }
    for(char c='a';c<='z';c++){
      cout<<c<<" ";
      pr_binary(int(c));
    }
    //difference between upper case letter and lower case letter binary is that
    //in upper case letter 5th bit!=1;
    //in lower case letter 5th bit =1;
    cout<<char('A'|(1<<5))<<endl;
    //in lower case;

    cout<<char('a'&(~(1<<5)))<<endl;
    //in upper case;

    //actually char of 1<<5 is _(space);
    //take any upper case letter and its ||  with space will get the corresponding lower case letter;

    cout<<char('C'|' ')<<endl;
   // will make it small c
//take any lower case letter and its || with _(underscore) will get the corresponding upper //case letter;`

    cout<<char('c'&'_')<<endl;
   // will make it capital C

 int a=4;
  int b=5;
  a=a^b;
  b=b^a;
  a=a^b;
  // cout<<a<<" "<<b;

   int n=16;
   n&(n-1)?dis("NO"):dis("YES");

   //Update : this will not work for n==0;
   for n=0;
   //we can have the function
   bool check_power_of_two(int num){
      return n && !(n&(n-1));
   }

     int i=4;
   //clearing upto 5 the place;
   int a=59;
   int b=(a&(~((1<<(i+1))-1)));
   //clearing the lsb upto ith bit;
   pr_binary(b);
   
   i=3;
   int c=(a&((1<<(i+1))-1));
   //clearing the msb upto ith bit;
   pr_binary(c);