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;
using ll = long long;
ll B[100001];
vector<int> max_bs[61];
vector<int> has_bit[61];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> B[i];
    int max_b = 0;
    for (int j = 0; j <= 60; j++) {
      if (B[i] & (1LL << j)) max_b = j;
      if (has_bit[j].empty()) has_bit[j].push_back(0);
    }
    max_bs[max_b].push_back(i);
  }
  bool w = 1;
  for (int max_b = 60; max_b >= 0; max_b--) {
    if (int(max_bs[max_b].size()) > int(has_bit[max_b].size())) {
      w = 0;
      break;
    }
    for (int i = 0; i < int(max_bs[max_b].size()); i++) {
      int cur_index = max_bs[max_b][i];
      for (int j = 0; j <= 60; j++) {
        if (B[cur_index] & (1LL << j)) has_bit[j].push_back(cur_index);
      }
    }
  }
  cout << (w ? "Yes" : "No") << "\n";
  if (!w) return 0;
  vector<int> ans;
  ll cur;
  int cur_max_b;
  for (int b = 0; b <= 60; b++) {
    if (!max_bs[b].empty()) {
      cout << B[max_bs[b].back()] << " ";
      max_bs[b].pop_back();
      cur = B[max_bs[b][0]];
      cur_max_b = b;
      break;
    }
  }
  while (1) {
    bool found = 0;
    for (int b = 0; b <= cur_max_b; b++) {
      if (!max_bs[b].empty() && !((1LL << b) & cur)) {
        found = 1;
        cur ^= B[max_bs[b].back()];
        cout << B[max_bs[b].back()] << " ";
        max_bs[b].pop_back();
        break;
      }
    }
    if (!found) {
      cur_max_b++;
      while (cur_max_b <= 60 && max_bs[cur_max_b].empty()) cur_max_b++;
      if (cur_max_b > 60) break;
      cur ^= B[max_bs[cur_max_b].back()];
      cout << B[max_bs[cur_max_b].back()] << " ";
      max_bs[cur_max_b].pop_back();
    }
  }
  cout << "\n";
}
