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;
long long int powmod(long long int a, long long int e) {
  if (e == 0) return 1;
  long long int x = powmod(a * a % 1000000007, e >> 1);
  return e & 1 ? x * a % 1000000007 : x;
}
long long int b[(long long int)100001];
multiset<long long int> st[60];
vector<long long int> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> b[i];
    long long int bit;
    for (bit = 59; bit >= 0; bit--)
      if ((1ll << bit) & b[i]) break;
    st[bit].insert(b[i]);
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    long long int mini = (1ll << 60);
    long long int flag = 0;
    long long int ansbit;
    for (long long int bit = 59; bit >= 0; bit--) {
      if (st[bit].empty()) continue;
      if (!(ans & (1ll << bit))) {
        flag = 1;
        if (*st[bit].begin() < mini) {
          ansbit = bit;
          mini = *st[bit].begin();
        }
      }
    }
    if (flag == 0) {
      cout << "No";
      return 0;
    }
    st[ansbit].erase(st[ansbit].begin());
    ans = ans xor mini;
    v.push_back(mini);
  }
  cout << "Yes\n";
  for (long long int i = 0; i < n; i++) cout << v[i] << " ";
}
