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;
template <typename T>
void debug_out(T t) {
  cerr << t;
}
template <typename A, typename B>
void debug_out(pair<A, B> u) {
  cerr << "(" << u.first << " " << u.second << ")";
}
template <typename T>
void debug_out(vector<T> t) {
  int sz = t.size();
  for (int i = 0; i < sz; i++) {
    debug_out(t[i]);
    if (i != sz - 1) cerr << ", ";
  }
}
template <typename T>
void debug_out(vector<vector<T>> t) {
  int sz = t.size();
  for (int i = 0; i < sz; i++) {
    debug_out(t[i]);
    if (i != sz - 1) cerr << endl;
  }
}
int n;
const int N = 60;
int main() {
  cin >> n;
  vector<long long> b(n);
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]);
  long long now = 1LL << N;
  set<pair<long long, long long>> cur;
  for (int i = N - 1; i > -1; i--) {
    vector<long long> bar;
    for (int j = 0; j < n; j++)
      if (b[j] & (1LL << i) && b[j] < (1LL << (i + 1))) bar.push_back(b[j]);
    vector<long long> id;
    for (auto j : cur)
      if (j.second & (1LL << i)) {
        id.push_back(j.first);
      }
    if (bar.size() > id.size() + 1) {
      cout << "No";
      return 0;
    }
    reverse(id.begin(), id.end());
    while (bar.size() && id.size()) {
      cur.insert(make_pair(id.back() - now, bar.back()));
      bar.pop_back();
      id.pop_back();
    }
    if (bar.size()) {
      cur.insert(make_pair((cur.size() ? prev(cur.end())->first : 0) + now,
                           bar.back()));
    }
    now >>= 1;
  }
  cout << "Yes\n";
  for (auto i : cur) {
    printf("%lld ", i.second);
  }
  return 0;
}
