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;
void sync(bool b) {
  cin.sync_with_stdio(b);
  cout.sync_with_stdio(b);
}
const int MAXN = 100005;
const int MAXBIT = 60;
long long b[MAXN];
vector<long long> groups[MAXBIT];
set<pair<long long, long long> > taked[MAXBIT];
void no() {
  cout << "No\n";
  exit(0);
}
long long getRand() {
  long long r = ((long long)rand() << 15) + rand();
  return r;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    int bit = MAXBIT;
    while ((b[i] & (1ll << bit)) == 0) bit--;
    groups[bit].push_back(b[i]);
  }
  set<pair<long long, long long> > ans;
  ans.insert(make_pair(-1e18, 0));
  ans.insert(make_pair(1e18, 1ll << MAXBIT));
  for (int i = MAXBIT - 1; i >= 0; --i) {
    if (groups[i].empty()) continue;
    if (groups[i].size() > taked[i].size() + 1) no();
    int index = 0;
    for (auto &p : taked[i]) {
      auto iter = ans.lower_bound(p);
      iter--;
      long long prevId = iter->first;
      long long id = (prevId + p.first) / 2;
      long long value = groups[i][index];
      ans.insert(make_pair(id, value));
      for (int j = 0; j < i; ++j)
        if (value & (1ll << j)) taked[j].insert(make_pair(id, value));
      index++;
      if (index == groups[i].size()) break;
    }
    if (index < groups[i].size()) {
      long long value = groups[i][index];
      long long lastId = taked[i].empty() ? ans.begin()->first - getRand()
                                          : taked[i].rbegin()->first;
      long long nextId = taked[i].empty()
                             ? ans.rbegin()->first + getRand()
                             : ans.upper_bound(*taked[i].rbegin())->first;
      long long id = (lastId + nextId) / 2;
      ans.insert(make_pair(id, value));
      for (int j = 0; j < i; ++j)
        if (value & (1ll << j)) taked[j].insert(make_pair(id, value));
    }
  }
  cout << "Yes\n";
  for (auto &p : ans)
    if (p.second > 0 && p.second < (1ll << MAXBIT)) cout << p.second << ' ';
  cout << '\n';
}
int main() {
  sync(false);
  solve();
  return 0;
}
