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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using std::abs;
using std::cerr;
using std::cin;
using std::cout;
using std::map;
using std::max;
using std::min;
using std::pair;
using std::set;
using std::string;
using std::swap;
using std::vector;
using ll = long long;
using uint = unsigned int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
void _dbg(const char* _s, T _h) {
  cerr << _s << " = " << _h << "\n";
}
template <typename T, typename... Ts>
void _dbg(const char* _s, T _h, Ts... _t) {
  int _b = 0;
  while (((_b += *_s == '(') -= *_s == ')') != 0 || *_s != ',') cerr << *_s++;
  cerr << " = " << _h << ",";
  _dbg(_s + 1, _t...);
}
struct init {
  init() {
    cin.tie(0);
    std::iostream::sync_with_stdio(0);
    cout << std::fixed << std::setprecision(10);
    cerr << std::fixed << std::setprecision(5);
    using namespace std::chrono;
    srand(duration_cast<nanoseconds>(
              high_resolution_clock::now().time_since_epoch())
              .count());
  }
  ~init() {}
} init;
const int MAXN = 1e5 + 15, B = 60;
ll b[MAXN];
ll r[MAXN];
void no() {
  cout << "No\n";
  exit(0);
}
vector<ll> ans[B + 1];
bool used[MAXN];
int main() {
  int n = rand() % 4 + 1;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    b[i] = r[i - 1] ^ r[i];
    ;
    cin >> b[i];
  }
  for (int bit = B - 1; bit >= 0; --bit) {
    vector<ll> xex;
    for (int i = 1; i <= n; ++i)
      if (b[i] >> bit & 1LL) {
        if (!used[i]) {
          used[i] = 1;
          xex.emplace_back(b[i]);
        }
      }
    uint t = 0;
    if (xex.size()) ans[bit].emplace_back(xex[t++]);
    for (ll i : ans[bit + 1]) {
      ans[bit].emplace_back(i);
      if ((i >> bit & 1LL) && t < xex.size()) ans[bit].emplace_back(xex[t++]);
    }
  }
  ll w = 0;
  ll a = 0;
  if (ans[0].size() != n) no();
  for (ll i : ans[0]) {
    a ^= i;
    if (a <= w) {
      no();
    }
    w = a;
  }
  cout << "Yes\n";
  for (ll i : ans[0]) {
    cout << i << ' ';
  }
  cout << '\n';
  return 0;
}
