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 long long mod = 1000000009;
const long long INF = 1000000010;
const long long LONGINF = 4e18;
const double PI = 3.141592653589793;
long long a[100100];
int st[100100];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<int> cur;
  vector<int> ones;
  for (long long d = 59; d >= 0; --d) {
    vector<int> bad;
    for (int i = 0; i < n; ++i) {
      if (st[i] == 0 && (a[i] & (1ll << d))) {
        bad.push_back(i);
        st[i] = 1;
      }
    }
    if (bad.size() > ones.size() + 1) {
      cout << "No" << endl;
      return;
    }
    vector<int> ones1, cur1;
    int curi = 0;
    for (int i = 0; i < bad.size(); ++i) {
      if (d > 0 && (a[bad[i]] & (1ll << (d - 1)))) {
        ones1.push_back(cur1.size());
      }
      cur1.push_back(bad[i]);
      if (i + 1 < bad.size()) {
        while (curi <= ones[i]) {
          if (d > 0 && (a[cur[curi]] & (1ll << (d - 1))))
            ones1.push_back(cur1.size());
          cur1.push_back(cur[curi++]);
        }
      }
    }
    while (curi < cur.size()) {
      if (d > 0 && (a[cur[curi]] & (1ll << (d - 1))))
        ones1.push_back(cur1.size());
      cur1.push_back(cur[curi++]);
    }
    cur = cur1;
    ones = ones1;
  }
  cout << "Yes" << endl;
  for (int i = 0; i < n; ++i) {
    cout << a[cur[i]] << ' ';
  }
  cout << endl;
}
void init() {}
int main() {
  iostream::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(20) << fixed;
  init();
  solve();
  return 0;
}
