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;
const int MOD = 1000000007;
void yes() {
  cout << "Yes"
       << "\n";
}
void no() {
  cout << "No"
       << "\n";
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
struct FAIL {
  FAIL() {
    cout << "CHANGE!!!"
         << "\n";
  }
};
vector<long long> my_merge(const vector<long long> &has,
                           const vector<long long> &ans) {
  if (ans.size() == 0) {
    if (has.size() >= 2) {
      cout << "No";
      exit(0);
    }
    return has;
  }
  long long sum = 0;
  vector<long long> res;
  int j = 0;
  int n = has.size();
  int m = ans.size();
  for (int i = (0); i < (n); ++i) {
    while (j < m && (sum >= (sum ^ has[i]))) {
      sum ^= ans[j];
      res.push_back(ans[j]);
      ++j;
    }
    if (sum >= (sum ^ has[i])) {
      cout << "No";
      exit(0);
    }
    res.push_back(has[i]);
    sum ^= has[i];
  }
  while (j < m) {
    sum ^= ans[j];
    res.push_back(ans[j]);
    ++j;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int __i = 0; __i < (n); ++__i) cin >> v[__i];
  ;
  vector<int> used(n);
  vector<long long> ans;
  for (int j = (61); j > (-1); --j) {
    vector<long long> has;
    for (int i = (0); i < (n); ++i) {
      if (!used[i] && (v[i] & (1LL << j))) {
        has.push_back(v[i]);
        used[i] = 1;
      }
    }
    ans = my_merge(has, ans);
  }
  assert(ans.size() == n);
  vector<long long> b(n);
  long long sum = 0;
  for (int i = (0); i < (n); ++i) {
    sum ^= ans[i];
    b[i] = sum;
  }
  assert(is_sorted(b.begin(), b.end()));
  cout << "Yes"
       << "\n";
  for (auto elem : ans) {
    cout << elem << " ";
  }
}
