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;
vector<int> gr[100000 + 5];
vector<int> g;
struct dat {
  long long val;
  dat* next;
} * a, *act;
dat* rev[100000 + 5];
long long imax, val[100000 + 5];
int viz[100000 + 5];
void group(long long val, int ind) {
  long long p, i;
  p = 1;
  i = 0;
  while (p <= val) {
    if ((val & p) != 0) {
      gr[i].push_back(ind);
    }
    p = p * 2;
    i++;
  }
  imax = max(imax, i - 1);
}
void adaug(int ind, dat* dupa) {
  act = new dat;
  act->val = ind;
  act->next = dupa->next;
  dupa->next = act;
  rev[ind] = act;
}
int main() {
  long long n, i, nrnev, ind, nrviz, siz;
  cin >> n;
  imax = -1;
  for (i = 1; i <= n; i++) {
    cin >> val[i];
    group(val[i], i);
  }
  if (gr[imax].size() > 1) {
    cout << "No" << '\n';
    return 0;
  }
  a = new dat;
  rev[gr[imax][0]] = a;
  viz[gr[imax][0]] = 1;
  a->val = gr[imax][0];
  a->next = 0;
  act = new dat;
  act->val = 0;
  act->next = a;
  a = act;
  imax--;
  while (imax >= 0) {
    nrnev = 0;
    nrviz = 0;
    g.clear();
    for (i = 0; i < gr[imax].size(); i++) {
      ind = gr[imax][i];
      if (viz[ind] == 0)
        g.push_back(ind);
      else {
        viz[ind] = imax + 1;
        nrviz++;
      }
    }
    nrnev = g.size();
    if (nrnev > nrviz + 1) {
      cout << "No" << '\n';
      return 0;
    }
    if (nrnev > 0) {
      ind = g[g.size() - 1];
      viz[ind] = -1;
      adaug(ind, a);
      g.pop_back();
      if (nrnev > 1) {
        siz = 0;
        act = a->next->next;
        while (!g.empty()) {
          while (viz[act->val] != imax + 1) {
            act = act->next;
          }
          viz[g[g.size() - 1]] = -1;
          adaug(g[g.size() - 1], act);
          g.pop_back();
          act = act->next;
        }
      }
    }
    imax--;
  }
  cout << "Yes" << '\n';
  act = a->next;
  while (act != 0) {
    cout << val[act->val] << ' ';
    act = act->next;
  }
  cout << '\n';
  return 0;
}
