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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
using namespace std;
const long long maxn = 5e5 + 123, MX = 1e9, inf = 1e9, block = 2;
long long n, x;
vector<long long> v[100];
int getb(long long x) {
  int res = 0;
  while (x != 0) x /= 2, res++;
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    v[getb(x)].push_back(x);
  }
  vector<long long> ans, tmp;
  for (int b = 62; b >= 0; b--) {
    long long cur = 0, j = 0;
    for (int i = 0; i < v[b].size(); i++) {
      while ((cur ^ v[b][i]) <= cur) {
        if (j == ans.size()) return cout << "No", 0;
        tmp.push_back(ans[j]);
        cur ^= ans[j];
        j++;
      }
      tmp.push_back(v[b][i]);
      cur ^= v[b][i];
    }
    while (j < ans.size()) tmp.push_back(ans[j++]);
    ans = tmp;
    tmp.clear();
  }
  cout << "Yes" << endl;
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
}
