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;
int clz(unsigned long long N) {
  return N ? 64 - __builtin_clzll(N) : -200000000;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  vector<long long> b(n);
  for (int i = 0; i < (n); i++) cin >> b[i];
  sort(b.begin(), b.end());
  vector<vector<long long> > ra(70, vector<long long>());
  for (int i = 0; i < (n); i++) {
    int r = clz(b[i]);
    ra[r].push_back(b[i]);
  }
  int cR = 1;
  long long curr = 0;
  while (ra[cR].size() == 0) {
    cR++;
  }
  vector<long long> ans;
  curr = ra[cR][0];
  ans.push_back(ra[cR][0]);
  ra[cR].erase(ra[cR].begin());
  while (ans.size() < n) {
    bool f = 0;
    long long t = curr;
    int i = 1;
    while (t) {
      while (t % 2 == 1) {
        t = t >> 1;
        i++;
      }
      if (ra[i].size()) {
        ans.push_back(ra[i][0]);
        curr = curr ^ ra[i][0];
        ra[i].erase(ra[i].begin());
        f = 1;
        cR = max(cR, i);
        break;
      } else {
        t >>= 1;
        i++;
      }
    }
    if (f) {
    } else {
      cR++;
      while (cR <= 64 && ra[cR].size() == 0) cR++;
      if (cR == 65) {
        break;
      } else {
        ans.push_back(ra[cR][0]);
        curr = curr ^ ra[cR][0];
        ra[cR].erase(ra[cR].begin());
      }
    }
  }
  if (ans.size() == n) {
    cout << "Yes\n";
    for (int i = 0; i < (n); i++) cout << ans[i] << " ";
  } else
    cout << "No";
}
