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;
int n;
priority_queue<long long> v[64];
int findSize(long long a) {
  int ret = 0;
  while (a != 0) {
    a /= 2;
    ret++;
  }
  return ret;
}
int first0(long long a) {
  int size = findSize(a);
  for (int i = size - 1; i >= 0; i--) {
    if ((((long long)1 << i) & a) == 0) return i + 1;
  }
  return 0;
}
vector<long long> res;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long imsi;
    cin >> imsi;
    v[findSize(imsi)].push(imsi);
  }
  long long sum;
  for (int i = 0; i < 64; i++) {
    if (!v[i].empty()) {
      sum = v[i].top();
      v[i].pop();
      res.push_back(sum);
      break;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int size = findSize(sum);
    int first = first0(sum);
    int k;
    for (k = 0; k < size; k++) {
      if ((((long long int)1 << k) & sum) == 0) {
        if (!v[k + 1].empty()) {
          sum ^= v[k + 1].top();
          res.push_back(v[k + 1].top());
          v[k + 1].pop();
          break;
        }
      }
    }
    if (k != size) continue;
    int j;
    for (j = size + 1; j < 64; j++) {
      if (!v[j].empty()) {
        sum ^= v[j].top();
        res.push_back(v[j].top());
        v[j].pop();
        break;
      }
    }
    if (j == 64) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] << ' ';
  }
}
