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;
vector<long long> v, v2, v3;
int main() {
  long long n, cl, ck, u, i, j;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 0; i < n; ++i) {
    cin >> j;
    v.push_back(j);
  }
  sort(v.begin(), v.end());
  j = n - 1;
  while (j >= 0) {
    long long t = 1;
    while (t <= v[j]) t *= 2;
    t /= 2;
    vector<long long> v1;
    while (j >= 0 && v[j] >= t) {
      v1.push_back(v[j]);
      --j;
    }
    sort(v1.begin(), v1.end());
    int j1 = 0;
    v2.clear();
    for (i = 0; i < v3.size(); ++i) {
      if ((v3[i] & t) > 0 && j1 < v1.size()) {
        v2.push_back(v1[j1]);
        ++j1;
      }
      v2.push_back(v3[i]);
    }
    if (j1 == v1.size() - 1) {
      v2.push_back(v1[j1]);
      ++j1;
    }
    if (j1 != v1.size()) {
      cout << "No";
      return 0;
    }
    v3 = v2;
  }
  cout << "Yes\n";
  for (i = 0; i < v2.size(); ++i) cout << v3[i] << ' ';
}
