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 std::cerr;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::fixed;
using std::ios_base;
using std::list;
using std::make_pair;
using std::make_tuple;
using std::map;
using std::nth_element;
using std::pair;
using std::queue;
using std::reverse;
using std::set;
using std::sort;
using std::stack;
using std::string;
using std::tuple;
using std::unordered_map;
using std::unordered_set;
using std::vector;
const int MAXN = 100000;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int64_t L_INF = 4e18;
int n;
vector<int64_t> numbers[64];
list<int64_t> ans;
int get_top(int64_t num) {
  for (int i = 62; i >= 0; i--) {
    if (num & (1ll << i)) {
      return i;
    }
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int64_t b;
    cin >> b;
    numbers[get_top(b)].push_back(b);
  }
  for (int i = 62; i >= 0; i--) {
    if (numbers[i].empty()) {
      continue;
    }
    if (numbers[i].size() > ans.size() + 1) {
      cout << "No\n";
      exit(0);
    }
    auto iter = ans.begin();
    iter = ans.insert(iter, numbers[i][0]);
    iter++;
    int j = 1;
    int64_t sum = numbers[i][0];
    while (j < (int)numbers[i].size() && iter != ans.end()) {
      sum ^= *iter;
      iter++;
      if ((sum ^ numbers[i][j]) > sum) {
        iter = ans.insert(iter, numbers[i][j]);
        sum ^= *iter;
        iter++;
        j++;
      }
    }
    if (j != (int)numbers[i].size()) {
      cout << "No\n";
      exit(0);
    }
  }
  cout << "Yes\n";
  for (int64_t b : ans) {
    cout << b << ' ';
  }
  return 0;
}
