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("O3")
#pragma GCC target("sse,sse2,sse3,sse4")
using namespace std;
const long long N = 2e5 + 10;
const long long B = 1700;
const long long N2 = 3e5 + 10;
const long long off = 1e3;
const long long Sinf = 1e9 + 10;
const long long Binf = 8e18 + 10;
const long long mod = 1e9 + 7;
const double eps = 1e-12;
const double pi = 3.14159265358979323846;
const double e = exp(1);
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
using namespace std;
long long n;
long long a[N];
vector<long long> mxbit[65];
long long getmxbit(long long first) {
  for (long long i = 61; i >= 0; --i)
    if ((first >> i) & 1ll) return i;
  return 0;
}
list<long long> solve(long long bit) {
  if (bit == 61) return list<long long>();
  list<long long> lst = move(solve(bit + 1));
  long long xr = 0;
  if (!mxbit[bit].empty()) {
    lst.insert(lst.begin(), mxbit[bit].back());
    mxbit[bit].pop_back();
  }
  for (auto it = lst.begin(); it != lst.end(); ++it) {
    long long val = *it;
    xr ^= val;
    if (mxbit[bit].empty()) break;
    if ((xr ^ mxbit[bit].back()) > xr) {
      ++it;
      it = lst.insert(it, mxbit[bit].back());
      xr ^= mxbit[bit].back();
      mxbit[bit].pop_back();
    }
  }
  if (!mxbit[bit].empty() && (xr ^ mxbit[bit].back()) > xr) {
    lst.insert(lst.end(), mxbit[bit].back());
    mxbit[bit].pop_back();
  }
  if (!mxbit[bit].empty()) {
    cout << "No"
         << "\n";
    exit(0);
  }
  return lst;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n; ++i) mxbit[getmxbit(a[i])].push_back(a[i]);
  auto lst = solve(0);
  cout << "Yes"
       << "\n";
  for (auto it : lst) cout << it << " ";
  cout << "\n";
}
