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 <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T sqr(T a) {
  return a * a;
}
template <class T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
const double pi = acos(-1.0);
int n;
int m;
int main() {
  cin >> n;
  vector<long long> a[64];
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    int len = 64 - __builtin_clzll(x);
    a[len].push_back(x);
  }
  vector<long long> cert;
  for (int i = 64; i >= 0; i--) {
    auto& cur = a[i];
    if (((int)(cur).size()) > 0) {
      long long mask = 0;
      int cnt = 1;
      vector<long long> new_cert;
      mask ^= cur.back();
      new_cert.push_back(cur.back());
      cur.pop_back();
      for (int j = 0; j < ((int)(cert).size()); j++) {
        mask ^= cert[j];
        new_cert.push_back(cert[j]);
        if ((mask & (1ll << (i - 1))) == 0) {
          if (((int)(cur).size())) {
            mask ^= cur.back();
            new_cert.push_back(cur.back());
            cur.pop_back();
          }
          cnt++;
        }
      }
      if (((int)(cur).size())) {
        puts("No");
        return 0;
      }
      cert = new_cert;
    }
  }
  cout << "Yes" << endl;
  for (auto x : cert) {
    cout << x << " ";
  }
  return 0;
}
