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;
const int INFL = (int)1e9;
const long long int INFLL = (long long int)1e18;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
int main() {
  int n;
  cin >> n;
  map<long long int, int> b;
  for (int i = 0; i < (int)n; i++) {
    long long int x;
    cin >> x;
    b[x]++;
  }
  long long int a = 0;
  vector<long long int> ans;
  for (int unused = 0; unused < (int)n; unused++) {
    bool judge = false;
    for (int i = 0; i < (int)62; i++) {
      if ((a >> i) % 2 == 0) {
        decltype(b)::iterator itr = b.lower_bound(1LL << i);
        if (itr == b.end()) continue;
        if (itr->first < (1LL << (i + 1))) {
          a ^= itr->first;
          ans.push_back(itr->first);
          itr->second--;
          if (itr->second == 0) b.erase(itr);
          judge = true;
          break;
        }
      }
    }
    if (!judge) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  for (int i = 0; i < (int)ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  ;
}
