/* LittleElephantAndDouble (SRM 597, D2, 250 from Topcoder).
   James S. Plank
   Wed Dec  4 11:34:32 EST 2019
   This is an O(n) solution that divides each element by two until it is no longer
   divisible by two.  It checks each element with A[0]. 
 */

#include <string>
#include <vector>
#include <cstdlib>
using namespace std;

class LittleElephantAndDouble {
  public:
    string getAnswer(vector <int> &A);
};

string LittleElephantAndDouble::getAnswer(vector <int> &A)
{
  size_t i;

  for (i = 0; i < A.size(); i++) {
    while ((A[i] & 0x1) == 0) A[i] >>= 1;   /* I'm using bit arithmetic instead of div and mod. */
    if (A[i] != A[0]) return "NO";
  }
  return "YES";
}