/* LittleElephantAndDouble (SRM 597, D2, 250 from Topcoder).
   James S. Plank
   Wed Dec  4 11:33:40 EST 2019
   This is an O(n log n) solution that sorts the vector, and then
   checks every element with its neighbor.
 */

#include <string>
#include <vector>
#include <list>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

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

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

  sort(A.begin(), A.end());

  /* Now, for each element, keep doubling it until it is greater than
     or equal to the next element.  If greater, return "NO".  If you 
     succeed for all elements, return "YES". */

  for (i = 0; i < A.size()-1; i++) {
    while (A[i] < A[i+1]) A[i] *= 2;
    if (A[i] != A[i+1]) return "NO";
  }
  return "YES";
}