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 n;
vector<long long> worki[61];
vector<long long> wszystkie;
vector<long long> perm;
vector<long long> nowy;
void wczytaj() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long a;
    cin >> a;
    wszystkie.push_back(a);
  }
}
inline long long pow(int k) {
  long long a = 1;
  a <<= k;
  return a;
}
void posegreguj() {
  for (int i = 60 - 1; i >= 0; i--)
    for (__typeof(wszystkie.begin()) it = wszystkie.begin();
         it != wszystkie.end(); it++)
      if (*it >= pow(i)) {
        worki[i].push_back(*it);
        *it = -1;
      }
}
bool znajdz_perma() {
  for (int i = 60 - 1; i >= 0; i--) {
    bool akt = 0;
    nowy.clear();
    int j = 0;
    int s = perm.size();
    for (__typeof(worki[i].begin()) it = worki[i].begin(); it != worki[i].end();
         it++) {
      while (akt && j < s) {
        nowy.push_back(perm[j]);
        if (perm[j] & pow(i)) akt = 0;
        j++;
      }
      if (akt == 0) {
        nowy.push_back(*it);
        akt = 1;
      } else
        return 0;
    }
    while (j < s) nowy.push_back(perm[j++]);
    perm.clear();
    for (__typeof(nowy.begin()) it = nowy.begin(); it != nowy.end(); it++)
      perm.push_back(*it);
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  wczytaj();
  posegreguj();
  if (znajdz_perma()) {
    cout << "Yes\n";
    for (__typeof(perm.begin()) it = perm.begin(); it != perm.end(); it++)
      cout << *it << " ";
    cout << endl;
  } else
    cout << "No\n";
  return 0;
}
