Vitya has learned that the answer for The Ultimate Question of Life, the Universe, and Everything is not the integer 54 42, but an increasing integer sequence a_1, …, a_n. In order to not reveal the secret earlier than needed, Vitya encrypted the answer and obtained the sequence b_1, …, b_n using the following rules:

  * b_1 = a_1;
  * b_i = a_i ⊕ a_{i - 1} for all i from 2 to n, where x ⊕ y is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of x and y. 



It is easy to see that the original sequence can be obtained using the rule a_i = b_1 ⊕ … ⊕ b_i.

However, some time later Vitya discovered that the integers b_i in the cypher got shuffled, and it can happen that when decrypted using the rule mentioned above, it can produce a sequence that is not increasing. In order to save his reputation in the scientific community, Vasya decided to find some permutation of integers b_i so that the sequence a_i = b_1 ⊕ … ⊕ b_i is strictly increasing. Help him find such a permutation or determine that it is impossible.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5).

The second line contains n integers b_1, …, b_n (1 ≤ b_i < 2^{60}).

Output

If there are no valid permutations, print a single line containing "No".

Otherwise in the first line print the word "Yes", and in the second line print integers b'_1, …, b'_n — a valid permutation of integers b_i. The unordered multisets \\{b_1, …, b_n\} and \\{b'_1, …, b'_n\} should be equal, i. e. for each integer x the number of occurrences of x in the first multiset should be equal to the number of occurrences of x in the second multiset. Apart from this, the sequence a_i = b'_1 ⊕ … ⊕ b'_i should be strictly increasing.

If there are multiple answers, print any of them.

Examples

Input

3
1 2 3


Output

No


Input

6
4 7 7 12 31 61


Output

Yes
4 12 7 31 7 61 

Note

In the first example no permutation is valid.

In the second example the given answer lead to the sequence a_1 = 4, a_2 = 8, a_3 = 15, a_4 = 16, a_5 = 23, a_6 = 42.
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
class TaskE {
 public:
  void solve(istream& in, ostream& out) {
    int n;
    in >> n;
    vector<i64> b(n);
    vector<vector<i64>> q(60);
    for (int i = 0; i < n; ++i) {
      in >> b[i];
      int msb = 0;
      for (i64 x = b[i]; x; x >>= 1, ++msb)
        ;
      q[msb - 1].push_back(b[i]);
    }
    vector<i64> res;
    i64 cur = 0;
    for (int i = 0; i < n; ++i) {
      i64 x = 0;
      for (int j = 0; j < 60; ++j) {
        if (!q[j].empty() && (cur & (1LL << j)) == 0) {
          x = q[j].back();
          q[j].pop_back();
          break;
        }
      }
      if (x == 0) {
        out << "No\n";
        return;
      }
      cur ^= x;
      res.push_back(x);
    }
    out << "Yes\n";
    for (i64 x : res) {
      out << x << " ";
    }
    out << "\n";
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  TaskE solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
